Finding this type of limit can be seen in quite a few calculus problems. $$ \lim_{n\to\infty}n!^\frac1n $$ And also easily solved by Stirling approximation, or ect.
Then, what will happen when $n$ goes to zero? We cannot define a factorial of non-integer, so I'll use $x\Gamma(x)$ instead of $x!$. The values of these two are exactly the same when $x$ is natural number.
So, what would be the limit of this? $$ \lim_{x\to0}(x\Gamma(x))^\frac1x $$
Gamma function has many forms, and I'll use Weierstrass form like below instead of familiar integral form $\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}dt$. $$ \Gamma(z)=\frac{e^{-\gamma z}}z\prod_1^\infty\frac{e^\frac zn}{1+\frac zn} $$ Then, our $(x\Gamma(x))^\frac1x$ changes into $e^{-\gamma}\prod_1^\infty\frac{e^\frac1n}{\left(1+\frac xn\right)^\frac1x}$.
Since $\lim_{x\to0}\left(1+\frac xn\right)^\frac1x=e^\frac1n$, so...
$$ \lim_{x\to0}(x\Gamma(x))^\frac1x=\lim_{x\to0}e^{-\gamma}\prod_1^\infty\frac{e^\frac1n}{\left(1+\frac xn\right)^\frac1x}=e^{-\gamma} $$ ($\gamma$ is Euler-Mascheroni constant.)