Compute $\displaystyle \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}.$
let $\displaystyle a_n = \frac {n^n}{n!}$ and note that the root test will give the same result as the ratio test. that is, we will use the fact that:
$\displaystyle \lim |a_n|^{1/n} = \lim \left| \frac {a_{n + 1}}{a_n} \right|$
Now compute $\displaystyle \lim \left| \frac {a_{n + 1}}{a_n} \right|$
if i did not make a mistake, the answer should be $\displaystyle e$
Here is a way to do it. But it might be cheating, because it is basically Stirling approximation.
Instead compute the $\displaystyle \ln $ of this sequence,
$\displaystyle \ln \left( \frac{n^n}{n!} \right)^{1/n} = \ln n - \frac{1}{n}\ln (n!)$.
But, $\displaystyle \ln (n!) = \sum_{k=1}^n \ln k \approx \int_1^n \ln x dx = n\ln n - n + 1$ by Euler-Maclaurin formula, the remainder term will happen to go to zero.
Thus, we are left with $\displaystyle 1 - \frac{1}{n} \to 1$. Which means $\displaystyle \ln a_n \to 1 \implies a_n \to e$.
$\displaystyle y=\lim_{n \to {\infty}}\bigg(\frac{n^{n}}{n!}\bigg)^{\frac{1}{n} }$....so $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{1}{n}\cdot\ln\bigg(\frac{n^n}{n!}\b igg)$...$\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{1}{n}ln\bigg(\cdot\frac{n^{n}}{\int _0^{\infty}e^{-x}x^{n}dx}\bigg)$...using l'hopital's rule and much simplification we get $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{ln(n)n!+n!-\int_0^{\infty}e^{-x}x^n\ln(x)dx}{n!}$...using l'hopitals again gives us$\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{ln(n)e^{-n}n^n+(n-1)!+e^{-n}n^n-ln(n)e^{-n}n^n}{e^{-n}n^{n}}$..simplifying we get $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!+e^{-n}n^n}{e^{-n}n^n}$...now by splitting we get $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!}{e^{-n}n^n}+1=0+1$...so $\displaystyle ln(y)=1$...so $\displaystyle y=e$
It appears to me all you need to do is put a bound on $\displaystyle n!$. Of course, Stirling approximation is the best way to go, but that would be cheating here. You can use what PaulRS did here to get good estimates on $\displaystyle n!$ and use squeeze theorem.
$\displaystyle \left\{ {\frac{{n^n }}
{{n!}}} \right\}^{\frac{1}
{n}} = \exp \left\{ {\frac{1}
{n}\ln \frac{{n^n }}
{{n!}}} \right\} = \exp \left\{ {\frac{1}
{n}\ln \left( {\prod\limits_{k\, = \,1}^n {\frac{n}
{k}} } \right)} \right\} = \exp \left\{ { - \frac{1}
{n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}
{n}} } \right\}.$
As $\displaystyle n\to\infty,\,{\frac{1}
{n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}
{n}} }$ is a Riemann sum for $\displaystyle \int_0^1\ln x\,dx,$ which is a fairly known integral, it follows that $\displaystyle \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}=e.$