How do you find the limit as $x$ approaches infinity of $x! / a^x$ for some integer $a$ The limit as $n$ approaches infinity of the following functions diverges:


*

*$n! / a^n$

*$n! / n^a$
For any fixed integer $a$. How can I show this?
EDIT: The original question was the second bullet point; however I really meant to type the first bullet.
 A: Hint
By ratio test
$$\lim_{n\to+\infty}\frac{\frac{a^{n+1}}{(n+1)!}}{\frac{a^n}{n!}}=0$$
thus the series $\sum\frac{a^n}{n!}$ is convergent and
$$\lim_{n\to+\infty}\frac{a^n}{n!}=0$$
thus
$$\lim_{n\to+\infty}\frac{n!}{a^n}=+\infty$$
For the second, it is easy to prove by induction that
$n!\geq 2^{n-1}$ and
$$\frac{n!}{n^a}\geq \frac{2^{n-1}}{n^a}$$
as the exponential is faster than the polynomial, the limit is $+\infty$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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*

*$\ds{\large{n! \over a^{n}}:\ ?\quad \mbox{as}\ n \to \infty}$.
\begin{align}
{n! \over a^{n}} & \sim {\root{2\pi}n^{n + 1/2}\expo{-n} \over a^{n}} =
\root{2\pi}n^{1/2} \pars{n \over a\expo{}}^{n} \to \infty
\end{align}



*$\ds{\large{n! \over n^{a}}:\ ?\quad \mbox{as}\ n \to \infty}$.


\begin{align}
\mbox{Similarly,}\quad
{n! \over n^{a}} & \sim {\root{2\pi}n^{n + 1/2}\expo{-n} \over n^{a}} =
\root{2\pi}n^{1/2}n^{n - a + 1/2}\expo{-n} \to \infty
\end{align}
A: HINT:
For $x$ and $a$ integers with $x>a$ we can write
$$\frac{x!}{x^a}=\left(\frac{x}{x}\frac{x-1}{x}\frac{x-2}{x}\cdots \frac{x-a+1}{x}\right)(x-a)!$$

If $x$ is real and $a$ is an integer, then 
$$x!=\int_0^\infty t^{x}e^{-t}\,dt$$
Now, apply L'Hospital's Rule $a$ times.
A: For a fixed $a$;
$n!=n(n-1)(n-2)(n-3)\ldots (n-(a-1)(n-a)(n-(a+1))\ldots 3.2.1$
So we have $\dfrac{n!}{n^a}=\dfrac{n(n-1)(n-2)(n-3)\ldots (n-(a-1)(n-a)(n-(a+1))\ldots 3.2.1}{n^a}$
$=(1-\dfrac{1}{n})(1-\dfrac{2}{n})\ldots \times f(n)$  where $f(n)>1$ and the product $ \to \infty $as $n\to \infty$
A: Intuitively, both the numerator and denominator of $\frac {n!}{a^n}$ have $n$ factors.   As $n$ gets large, most of the factors in the numerator are much larger than $a$, while all the factors in the denominator are $a$, so the numerator grows faster and the fraction diverges.  
To be more rigorous, note that for $n=a^2$, you can pair the factors of $n!$ as $(1,a^2), (2,a^2-1), (3,a^2-2), \ldots (\frac 12a^2-1, \frac 12a^2)$ and the product of each pair is greater than $a^2$, so the whole product is greater than $(a^2)^{a^2/2}=a^{(a^2)}$ and the fraction is greater than $1$.  Each upward step in $n$ now multiplies the fraction by something greater than $\frac {a^2}a$, so the fraction diverges.  
You can also use Stirling's approximation for the factorial, that $n! \approx \frac {n^n}{e^n}\sqrt{2 \pi n}$ and find $\frac {n!}{a^n} \gt \frac {n^n}{(ae)^n}=\left(\frac n{ae} \right)^n$  which clearly diverges as $n$ gets greater than $ae$
