Is there an elementary way of proving $$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n,$$ given $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n,$$ without using L"Hopital's rule, Binomial Theorem, derivatives, or power series?
In other words, given the above restrictions, we want to show $$\left(\lim_{n\to\infty}\left(1+\frac1n\right)^n\right)^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n.$$
If you accept that exponentiation is continuous, then certainly $$\left(\lim_{n\to\infty}\left(1+\frac1n\right)^n\right)^x = \lim_{n\to\infty}\left(1+\frac1n\right)^{nx}$$ But if $u=nx$, then by substitution we have $$ \lim_{n\to\infty}\left(1+\frac1n\right)^{nx}=\lim_{u\to\infty}\left(1+\frac{x}{u}\right)^u $$
I'm not convinced that the substitution argument in the answer above does anything that isn't simply sleight of hand - i.e. invoking a property of limits which may or may not be true in the circumstances at hand and so I am dubious of its value as an explanation of the identity $e^x = \lim(1+\frac{x}{n})^n$ without including more detail somewhere. The following is one attempt to do that:
Theorem: If we assume the following:
1). If $(a_n)_{n=0}^\infty$ is a sequence and $\lim_{n \to \infty} a_n =l$, then for any subsequence $(a_{n_k})_{k=0}^{\infty}$ we have $\lim_{k \to \infty} a_{n_k} = \ell$ also.
2). $\lim_{n \to \infty} (1+\frac{x}{n})^n$ exists for all $x \in \mathbb R$.
then it follows that for $x \in \mathbb Q$ we have $\lim_{n \to \infty} (1+\frac{x}{n})^n =e^x$.
Proof: First suppose that $x =1/q$ for $q \in \mathbb Z_{>0}$. By 1) we have $$ e=\lim_{n \to \infty} (1+\frac{1}{n})^n = \lim_{m\to \infty} (1+\frac{1}{mq})^{mq} = \lim_{m \to \infty} \left((1+\frac{1}{mq})^m\right)^q $$ But now $t \mapsto t^q$ is continuous, so we see that $$ e = \left(\lim_{m\to \infty}\left(1+(\frac{q^{-1}}{m})\right)^m\right)^q \implies e^{q^{-1}} = \lim_{m\to \infty} \left(1+\frac{q^{-1}}{m}\right)^m $$ where the implication follows by taking the unique positive $q$-th root.
Now suppose that $x=pq^{-1}\in \mathbb Q_{>0}$. Then by assumption 2), $\lim_{m\to \infty}(1+\frac{x}{m})^m$ exists, and hence its limit is equal to the limit of the subsequence of where $m=pk$, with $k$ an integer: $$ \lim_{m \to \infty}(1+\frac{pq^{-1}}{m})^m = \lim_{pk \to \infty} (1+\frac{pq^{-1}}{pk})^{pk} = \lim_{k\to \infty} (1+\frac{q^{-1}}{k})^{pk} = \lim_{k \to \infty}\left((1+\frac{q^{-1}}{k})^k\right)^p $$ Now using continuity of $x \mapsto x^p$ it follows that $$ \lim_{m \to \infty} (1+\frac{pq^{-1}}{m})^m = \left(\lim_{k \to \infty}(1+\frac{q^{-1}}{k})^k\right)^p = (e^{q^{-1}})^p = e^{p/q}. $$ Thus we see that for $x \in \mathbb Q_{>0}$, provided that the limits $(1+\frac{x}{m})^m$ exist, then they are equal to $e^x$.
Finally, for the case where $x\in \mathbb Q_{<0}$, note that $$ (1+\frac{x}{n})^n(1-\frac{x}{n})^n = (1-\frac{x^2}{n^2})^n $$ Now it is easy to see that $(1+t)^n \geq 1+nt$ for all $n \in \mathbb Z_{>0}$ and $t \geq -1$ (see below). Thus for $n>|x|$ we find $$ 1>(1-\frac{x^2}{n^2})^n \geq 1-\frac{x^2}{n} \to 1, $$ and hence $\lim_{n \to \infty}(1-\frac{x}{n})^n = 1/e^x=e^{-x}$.
[Aside: It is not very difficult to use the argument which shows that $(1-\frac{x}{n})^n \to e^{-x}$ as $n\to \infty$ to establish 2) by first show it is true for $x<0$.]
Proof of the inequality: For the inequality $(1+t)^n \geq 1+nt$, the case $n=1$ this is trivial, while if it holds for $n\in \mathbb Z_{>0}$ then $$ (1+t)^{n+1} = (1+t)(1+t)^n \geq (1+t)(1+nt) = 1+(n+1)t + nt^2 \geq 1+(n+1)t $$ where we use $1+t\geq 0$ in the first inequality and $nt^2\geq 0$ in the last one.
