TITLE: How to prove that $\lim\limits_{n \to \infty} \frac{k^n}{n!} = 0$
QUESTION [17 upvotes]: It recently came to my mind, how to prove that the factorial grows faster than the exponential, or that the linear grows faster than the logarithmic, etc...
I thought about writing:
$$
a(n) = \frac{k^n}{n!} = \frac{ k \times k \times \dots \times k}{1\times 2\times\dots\times n} = \frac k1 \times \frac k2 \times \dots \times \frac kn = \frac k1 \times \frac k2 \times \dots \times \frac kk \times \frac k{k+1} \times \dots \times \frac kn
$$
It's obvious that after k/k, every factor is smaller than 1, and by increasing n, k/n gets closer to 0, like if we had $\lim_{n \to \infty} (k/n) = 0$, for any constant $k$.
But, I think this is not a clear proof... so any hint is accepted.
Thank you for consideration.

REPLY [4 votes]: As @robjohn has mentioned, this question is indeed very similar to Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$. the only question is who to play the role of $7$ in our case ($7$ is the smallest number for which $7!>(2+1)^7$ ). The general proof is the following: let us prove first that for any $j\in\mathbb N$ it holds that $n!\geq j^n$ for all $n$ large enough.

First, let us put $N = j^2+j$, then
$$
\frac{N!}{j^N} = \underbrace{\frac1j\cdot\frac2j\cdot\dots\cdot\frac jj}_{\pi_1}\cdot\underbrace{\frac{j+1}{j}\cdot\dots\cdot\frac{j^2}j}_{\pi_2}\cdot\underbrace{\frac{j^2+1}{j}\cdot\dots\cdot\frac{j^2+j}j}_{\pi_3}.
$$
Now, $\pi_2\geq 1$ since all terms in the product are grater than $1$. Moreover, by multiplying correspondent terms
$$
\pi_1\pi_3 = \frac{j^2+1}{j^2}\cdot\frac{2(j^2+2)}{j^2}\cdot\dots\cdot\frac{j(j^2+j)}{j^2}\geq1
$$
hence $N!\geq j^N$. 
For all $n\geq N$ it clearly holds now that $n!\geq j^n$. By induction: it holds for $n=N$. If it holds for some $n$ then
$$
(n+1)!\geq n!\cdot n\geq j^n\cdot n\geq j^{n+1}
$$
since $n\geq N\geq j$.

We proved that for any $j\in\mathbb N$ there is $N(j)$ s.t. $n\geq N(j)\Rightarrow n!\geq j^n$. Now let us put $j=[k]+1$, then
$$
0\leq\lim\limits_{n\to\infty}\frac{k^n}{n!}\leq \lim\limits_{n\to\infty}\left(\frac k{[k]+1}\right)^n = 0.
$$