Evaluate $\lim_{x\to \infty}\ (x!)^{1/x}$ Here's the problem...
$$\lim\limits_{x\to \infty}\ (x!)^{1/x}$$
I've deduced the answer to this is $\infty$, but haven't exactly shown that.
I'm getting $\infty^0$, so did both tricks where you change it to 
$$\exp\left({\lim_{x\to\infty}} \frac{1/x}{1/\ln(x!)}\right) = \exp\left(\frac{0}{0}\right)$$
And
$$\exp\left({\lim_{x\to\infty}} \frac{\ln(x!)}{1/(1/x)}\right) = \exp\left(\frac{\infty}{\infty}\right)$$
And for both of which, I need to do L'Hospital on this $\ln(x!)$ which I haven't the slightest idea how to do... Any suggestions that don't include Stirling's Approximation? My prof eluded to this being out of the question unless we were going to prove Stirling's first
 A: Hint:
For any fixed $a>0$, we eventually have $x!>a^x$, which means eventually $(x!)^{1/x}>(a^x)^{1/x}=a$.
A: $$A=(n!)^{\frac{1}{n}}\\ \ln(A)=\frac{1}{n}\ln(n!)$$
$$\lim_{n \rightarrow \infty} \ln(A)=\\\lim_{n \rightarrow \infty} \frac{1}{n}(ln(n!))=\\ \lim_{n \rightarrow \infty} \frac{ln(n)+ln(n-1)+ln(n-2)+....}{n}=\\ hop \\
\lim_{n \rightarrow \infty} \frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+...=\\ \lim_{n \rightarrow \infty}\sum_{k=2}^\infty \frac{1}{k}\to \infty \\ \ln(A) \to \infty \\A \to \infty$$
second way 
we know $$\color{red} {\lim_{n \rightarrow \infty}\frac{1}{n}(n!)^\frac{1}{n}=\frac{1}{e}} \to \\$$so  $$\lim_{n \rightarrow \infty}(n!)^\frac{1}{n}=\lim_{n \rightarrow \infty}\frac{n}{e}=\infty$$
A: It is easy to show that $n!>\left(\frac n2\right)^{n/2}$.  In fact, $(2n)!>n!\,n^n$.  
Therefore, given any number $B>0$, however large, we have
$$\begin{align}
\left(n!\right)^{1/n}&>\left(\frac n{2}\right)^{1/2}\\\\
&>B
\end{align}$$
whenever $n>2B^2$.  And we are done!
A: Let $y = (x!)^{\frac{1}{x}}$. Then, $\ln y = \frac{\ln x!}{x}$. Using Sterlings approximation of $\ln x! \sim x\ln x - x$ gives $\ln y \sim \ln x-1$ as $x \to \infty$. This means $y \to \infty$.
