Limit of $\lim \limits_{x \rightarrow \infty}\frac{(x!)^n}{(ax)!}$

For a given $n \geq 1$ $\hspace{2mm}(n \in \mathbb{R})$, I know that

$$\lim \limits_{x \rightarrow \infty}\frac{(x!)^n}{(ax)!},$$ only exists and it is equal to zero if $a \geq n.$ However, I cannot prove that.

Could you give me some ideas? Thanks in advance!

• Is $x$ a real number? – Siminore Jan 6 '15 at 9:30
• Yes! I have edited it. – Alex Silva Jan 6 '15 at 9:32
• Use Stirling's formula. – Yuval Filmus Jan 6 '15 at 9:34

You can use Stirling's formula: $x! \sim \sqrt{2\pi x} (x/e)^x$. In your case, assuming $a > 0$ you get $$\frac{(x!)^n}{(ax)!} \sim (2\pi)^{(n-1)/2} \frac{x^{nx}}{e^{nx}} \frac{e^{ax}}{(ax)^{ax}} = (2\pi)^{(n-1)/2} \frac{(x/e)^{(n-a)x}}{a^{ax}}.$$ If $a \geq n$ and $a > 1$ then this tends to zero. If $n > a$ then it tends to infinity. This leaves the corner case $a = n = 1$, in which the limit is $1$.