There is a way to define the natural numbers directly. Unlike the standard definition, it is not self-referential in nature. Before getting to that, I will review the standard definition. The most well-known construction of the natural numbers, found in texts such as those by Enderton or Hrbacek and Jech, begins as follows:
An inductive set is any set $X$ such that $\emptyset \in X$, and for all sets $a$, if $a \in X$ then $S(a) \in X$.
Here, $S(a) = a \cup \{a\}$ denotes the successor of a. Then the Axiom of Infinity states:
There exists an inductive set.
This definition is self-referential in the sense that the property "if $a \in X$ then $S(a) \in X$" references $X$ itself. Therefore, rather than defining the extension (members) of $X$ explicitly, as the Axiom Schema of Specification does for example, this property defines the extension implicitly, making it less clear what $X$ actually is.
The final step is to define the natural numbers $\mathbb N$ as the intersection of all inductive sets. This does result in a unique set, but it does not fully resolve the self-referential nature of the definition, in the sense that it still doesn't immediately describe the members of $\mathbb N$ and their properties.
There is nothing necessarily wrong with this, nor is there any formal circularity or logical contradiction here. Rather the issue is more of a meta-circularity, and has more to do with bringing the self-evidence of the Axiom of Infinity into question. None of the other axioms of ZFC use these kind of self-referential definitions.
One could argue that rigorous (e.g. ZFC) set theory should attempt to model the intuition of naive set theory, at least as far as it can while avoiding contradictions like Russell's Paradox. In naive set theory, sets are thought of as obtained by collecting together objects using some criterion. This is formalized in the Axiom Schema of Unrestricted Comprehension. While this axiom schema leads to paradoxes, these can be avoided using its restricted form, the Axiom Schema of Specification. The idea is to specify exactly what elements are in the defined set.
This is a unifying and intuitively appealing conceptual approach, which the standard definition of $\mathbb N$ does not follow. On the other hand, defining $\mathbb N$ as the smallest inductive set does immediately indicate that proof by induction is applicable. So that is perhaps one motivation for that definition.
But in the case of your question, it is precisely this self-referential aspect of the the Axiom of Infinity which I believe is the issue. Your attempt to define $\mathbb N$ as
$\mathbb N = \{x \mid x = S^n(\emptyset) \, \text{for some finite} \, n\}$.
will not work, because we cannot write $S^n$, the $n$-fold composition of the successor function $S$, in our first-order language, only in the meta-language. In fact, we can't write $f^n$ in our language for any function $f$ (set or class). We can only express it indirectly as an element of a set guaranteed to exist by the Axiom of Infinity and the Axiom Schema of Replacement. But then, this already requires a definition of $\mathbb N$.
However, we can define what it means to be finite directly. As I mentioned above, there is a direct (i.e. comprehension-like) way to define the natural numbers. We do this using ordinals.
If we take the common von Neumann definition of ordinals, then an ordinal is a transitive set that is well-ordered by set membership ($\in$). Furthermore, we define an ordinal $\alpha$ to be finite if every non-empty subset of $\alpha$ has a maximal element (with respect to the order defined by $\in$). Note the similarity of this definition to the well-founded property of a well-ordering, namely that every non-empty subset has a minimal element. From these definitions, it follows that the first few finite ordinals are $\emptyset$, $\{\emptyset\} = S(\emptyset)$, $\{\emptyset, \{\emptyset\}\} = S^2(\emptyset)$, etc.
Then we can define the natural numbers $\omega$ to be
$\omega = \{\alpha \mid \alpha \, \text{is a finite ordinal}\}$.
Of course, without an axiom asserting that such a set exists, this is just a class. We could just replace the Axiom of Infinity with the assertion that the above class is a set.
But we may as well just use the Axiom of Infinity, because as it turns out, using the definitions above, $\omega = \mathbb N$. In the remainder of this answer, I will prove this.
Proof. First we prove $\mathbb N \subseteq \omega$. We will show that $\omega$ is an inductive set. First note that $\emptyset \in \omega$. Now let $\alpha \in \omega$. Then $\alpha$ is a transitive set, and $\alpha \subseteq \alpha \cup \{\alpha\} = S(\alpha)$, so $S(\alpha)$ is a transitive set as well. Furthermore, since $\alpha$ is an ordinal, the set membership relation $\in$ defines a trichotomous, transitive, well-founded order on $\alpha$. Since for all $\beta \in S(\alpha)$, we have either $\beta \in \alpha$ or $\beta = \alpha$ by the definition of $S(\alpha)$, this order extends to a trichotomous, transitive order on $S(\alpha)$. To see that this order is also well-founded on $S(\alpha)$, let $A \subseteq S(\alpha)$ be non-empty. If $\alpha \notin A$ then $A \subseteq \alpha$, hence $A$ has a minimal element because $\alpha$ is an ordinal. Otherwise, by trichotomy of the order, the minimal element of $A$ is either $\alpha$ or the minimal element of $A \cap \alpha$, the latter of which exists because $A \cap \alpha \subseteq \alpha$. Finally, note that $\alpha$ is a maximal element of $S(\alpha) = \alpha \cup \{\alpha\}$, since $\alpha \not \in \alpha$, and for all $\beta \in \alpha$, $\alpha \notin \beta$ by trichotomy of the order. Therefore, every subset $A \subseteq S(\alpha)$ has a maximal element, which is either guaranteed to exist by the finiteness of $\alpha$ if $A \subseteq \alpha$, or is equal to $\alpha$ otherwise. Thus $S(\alpha)$ is a finite ordinal, so $S(\alpha) \in \omega$. This shows that $\omega$ is indeed an inductive set. Since $\mathbb N$ is the intersection of all inductive sets, $\mathbb N \subseteq \omega$.
Now we prove that $\omega \subseteq \mathbb N$. Suppose, for a contradiction, that there is an $\alpha \in \omega$ for which $\alpha \notin \mathbb N$. Since $\emptyset \in \mathbb N$ by definition, we may assume $\alpha \neq \emptyset$. Since $\alpha$ is a non-empty finite ordinal, it has a maximal element, call it $\gamma_0$. Then by the trichotomy of the set membership order on $\alpha$, for all $\beta \in \alpha$ either $\beta = \gamma_0$ or $\beta \in \gamma_0$. Therefore $\alpha = S(\gamma_0)$. Since $\alpha \notin \mathbb N$ and $\mathbb N$ is inductive, we must have $\gamma_0 \notin \mathbb N$. So the set $B = \{\beta \in \alpha \mid \beta \notin \mathbb N\}$ is non-empty. Since $B \subseteq \alpha$ and $\alpha$ is an ordinal, $B$ has a minimal element, call it $\beta_0$. Then $\beta_0$ must be non-empty, for otherwise $\beta_0 = \emptyset \in \mathbb N$ by definition of $\mathbb N$, contradicting the definition of $B$. Since $\beta_0 \in \alpha$ and $\alpha$ is a transitive set, $\beta_0 \subseteq \alpha$. Therefore, being a non-empty subset of the finite ordinal $\alpha$, $\beta_0$ has a maximal element, call it $\gamma_1$. By the same reasoning as above, $\beta_0 = S(\gamma_1)$. But then, because $\beta_0 \notin \mathbb N$ and $\mathbb N$ is inductive, we must also have $\gamma_1 \notin \mathbb N$. Note that $\gamma_1 \in \beta_0 \in \alpha$, so by transitivity $\gamma_1 \in \alpha$. This, together with the fact that $\gamma_1 \notin \mathbb N$, means that $\gamma_1 \in B$, contradicting the minimality of $\beta_0$. Therefore $\alpha$ must be an element of $\mathbb N$, and thus $\omega \subseteq \mathbb N$.
We conclude that $\omega = \mathbb N$. $\blacksquare$
