Limit of $(n!)^{1/(n!)}$ Wolfram gives me $\lim \limits_{n \to \infty} (n!)^{1/(n!)} = 1$
What is the analytical approach to compute this limit?
Thanks
 A: Let $x=n!$. Then the limit becomes 
\begin{align}
\lim_{x \rightarrow \infty} x^{1/x} &= \exp\left(\ln\left(\lim_{x \rightarrow \infty} x^{1/x}\right)\right) \\
&= \exp\left(\lim_{x \rightarrow \infty} \frac{1}{x}\ln x\right) \\
&= \exp(0) & \text{by L'Hopital's Rule} \\
&= 1
\end{align}
A: It’s the same as 
$$\begin{align*}
\lim_{x\to\infty}x^{1/x}&=\lim_{x\to\infty}e^{(\ln x)/x}\\
&=e^{\lim\limits_{x\to\infty}(\ln x)/x}\\
&=e^0\\
&=1\;.
\end{align*}$$
The second step is justified by the continuity of the exponential function.
A: METHOD 1:
There have already been a couple of quality answers posted.  But, I thought that it would be instructive to see a simple application of the Gamma function.  Recall $\Gamma(n+1)=n!$. Then, 
$$\begin{align}
\lim_{n\to \infty}\left(n!^{1/n!}\right)&=\lim_{n\to \infty}\left(e^{\frac{\log \Gamma(n+1)}{\Gamma(n+1)}}\right)\tag 1\\\\
&=e^{\lim_{n\to \infty}\left(\frac{\log \Gamma(n+1)}{\Gamma(n+1)}\right)}\tag 2\\\\
&=e^{\lim_{n\to \infty}\left(\frac{1}{\Gamma(n+1)}\right)}\tag 3\\\\
&=1\tag 4
\end{align}$$

NOTES:
In arriving at $(1)$, we used the definition of $\Gamma(n+1)=n!$ along with $x=e^{\log x}$ with $x=\Gamma(n+1)^{1/\Gamma(n+1)}$
To go from $(1)$ to (2)$, we used the continuity of the exponential to bring the limit inside the argument.
To go from $(2)$ to $(3)$, we used L'Hospital's Rule.
To go from $(3)$ to $(4)$, we observed again that $\Gamma(n+1) \to \infty$ as $n\to \infty$.

METHOD 2:
Here is another way to proceed without using L'Hospital's Rule. We have $a_n=n!^{1/n!}$.  Let $b_n=\log a_n$.  Then,
$$\begin{align}
b_n&=\frac{\log n!}{n!}\\\\
&=\frac{\sum_{k=1}^{n}\log k}{n!}\\\\
&=\frac{\frac1n\sum_{k=1}^{n}\log (k/n)+\log n}{(n-1)!}
\end{align}$$
Note that the sum is a Riemann sum for $\int_0^1 \log x dx=-1$.  Thus, the limit of $b_n$ is the same as $\lim_{n\to \infty}\frac{\log n}{(n-1)!}=0$.  And we recover the coveted limit 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}n!^{1/n!}=1}$$ 
