Currently, I have that $\sqrt n \leq \sqrt[n]{n!}$ for all integers $n≥1$, and since the sequence in the lower bound $\lim\limits_{n\to \infty}\sqrt n=\infty $ blows up, so does the upper bound $x_n=\sqrt[n]{n!}$
This feels quite hand-wavey and unsatisfying to me, so I was curious if there any other creative / more formal ways to prove its divergence?
This is an immediate consequence of the exponential series $\sum x^n/n!$ having infinite radius of convergence.
The radius of convergence $R$ of a real power series $\sum a_nx^n$ can be described by $1/R = \varlimsup \sqrt[n]{|a_n|}$. Taking $a_n = 1/n!$, so $R = \infty$ thanks to $e^x$ having an infinite radius of convergence, $0 = \varlimsup \sqrt[n]{1/n!} = \varlimsup 1/\sqrt[n]{n!}$. When a sequence of nonnegative numbers has $\varlimsup$ equal to $0$, the sequence tends to $0$, so $1/\sqrt[n]{n!} \to 0^+$. Thus $\sqrt[n]{n!} \to \infty$.
Alternatively, since $e^n = \sum_{k \geq 0} n^k/k!$, looking at the $n$-th term on the right shows $e^n > n^n/n!$ when $n \geq 1$, so $n! > n^n/e^n$. Now take $n$th roots: $\sqrt[n]{n!} > n/e$ when $n \geq 1$, so $\sqrt[n]{n!} \to \infty$ as $n \to \infty$.
$$\sqrt[n]{n!}= e^{1/n\cdot \ln n!}. $$
Using Stirling's approximation, we have $\ln n!\sim n\ln n-n,$ and so the limit of the exponent is $\lim_{n\to\infty}\frac {n\ln n-n}n=\lim_{n\to\infty}\ln n-1=\infty,$ and so the limit is infinite.
Assume that the sequence is bounded by a constant $M$. WLOG, let $M$ be an integer, then we have $$M^n > n!$$ is true for all $n$. However this is clearly false, since the LHS increases only by $M$ for each increment in $n$ while the $RHS$ increases by $n$, which can get arbitrarily large. Formally, let $c = \frac{M^M}{M!}$ and $k = \lceil \log_{(M+1)/M}c\rceil$. Then $$(M+k)!>M!(M+1)^k>M!M^k\left(\frac{M+1}{M}\right)^k>\frac{M^{M+k}}{c}c=M^{M+k}$$ Hence the assumption leads to contradiction for $n = M+k$. Q.E.D
