Prove that $a^nProve that $a^n <n! < n^n$ when $a>1$ and n always big enough.
How can I solve that? I need to prove that $$\lim\limits_{n\to\infty} \frac{a^n}{n!} = 0$$ and 
$$\lim\limits_{n\to\infty} \frac{n!}{n^n} = 0.$$

Is it possible to solve by another way? Some like this : 

 A: If you just want to calculate the limits, you could look at the series $$\sum\limits_{n=0}^\infty \frac{a^n}{n!}~\text{and}~\sum\limits_{n=1}^\infty \frac{n!}{n^n}$$ and check for convergence of these series.
A: Also, regarding the first Statement, I don't think this is actually true, consider a=3 and n=4, then $a^n$ = 81 > n! = 24.
A: Let's look at the ratio 
$$\frac{a^{n+1}/(n+1)!}{a^n/n!}=\frac{a}{n+1}<\frac12$$
for $n>2a-1$.  Thus, for $n>2a-1$, the terms of the sequence decay faster than the geometric progression $\frac{1}{2^n}$.  By comparison, the sequence tends to zero.

Now let's look at $\lim_{n\to \infty}\frac{n!}{n^n}$.  Recall Stirling's formula
$$n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac{1}{n}\right)\right)$$
Then, we have
$$\begin{align}
\lim_{n\to \infty}\frac{n!}{n^n}&=\lim_{n\to \infty}\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac{1}{n}\right)\right)}{n^n}  \\\\
&=\lim_{n\to \infty}\sqrt{2\pi n}e^{-n}\left(1+O\left(\frac{1}{n}\right)\right)\\\
&=0
\end{align}$$
A: $$0<\left| \frac { { a }^{ n } }{ n! }  \right| =\frac { \left| a \right|  }{ 1 } \frac { \left| a \right|  }{ 2 } \frac { \left| a \right|  }{ 3 } ...\frac { \left| a \right|  }{ m } \quad \frac { \left| a \right|  }{ (m+1) } ...\frac { \left| a \right|  }{ n } \le \frac { \left| { a }^{ m } \right|  }{ m! } { \left( \frac { \left| a \right|  }{ m+1 }  \right)  }^{ n-m }<\varepsilon \\ $$
for every $\forall \varepsilon >0$ and $m+1>\quad \left| a \right| $
A: The first limit is evaluable using the Stolz-Cesàro theorem. It is equal to $$\lim_{n\to\infty}\frac{a^{n+1}-a^n}{(n+1)!-n!}=\lim_{n\to\infty}\frac{a^{n}(a-1)}{n\cdot n!}=0,$$ where the rightmost equality follows from the upperboundedness of $\displaystyle\frac{a^n}{n!}$, which in turn is a consequence of its obvious decreasingness for large enough $n$.
As for the second one, we show by induction that for $n>1$ $$n!<n^n.\tag{1}$$ We have $$2=2!<2^2=4,$$and $$\frac{(n+1)!}{n!}<\frac{(n+1)^{n+1}}{n^n} \\ n+1<(n+1)\left(1+\frac{1}{n}\right)^n \\ \left(1+\frac{1}{n}\right)^n>1,$$ therefore $(1)$ does hold for $n>1$. If we suppose $$\lim_{n\to\infty}\frac{n!}{n^n}\ne0,$$ then from $$\min\left({\frac{n!}{n^n}}\right)\le\frac{n!}{n^n}<1 $$ follows $$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=1,$$ and again, by Stolz-Cesàro, $$\lim_{n\to\infty}\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}=1.$$ But combining $$\sqrt{2}-1<1 $$ and the decreasingness of $$\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}, $$ we get a contradiction.
