How to evaluate $\lim_{n\to\infty}n!/n^{k}$ I do not know how to go about finding the limit of this sequence. I know it diverges to infinity yet I can't find terms to use the squeeze lemma effectively.
$a_n = \frac{n!}{n^{1000}}$ 
$\lim_{n\rightarrow\infty} \frac{n!}{n^{1000}}$
I know that when $n>1000$, the numerator will "run out" of denominators and the expression will grow, yet I don't know how to formally prove this divergence.
In addition to this, since it is a sequence, no series laws can be applied. 
Anyone have an idea on how to approach this type of problem?

$$\log\frac{n!}{n^k} = \log n!-\log n^k = \sum_2^n \log i - k\log n$$
Let's apply the intergral test. Note that:
$$\int_2^{n} \log x\; dx-k \log n = x\log x\Bigg|_{2}^{n} - k\log n$$
So, for $n>k$, we get:
$$x\log x\Bigg|_{2}^{n} - k\log n = (n-k)\log -  2\log 2  n>0$$
And
$$\lim_{n\to \infty} (n-k)\log n - 2\log 2 = \infty$$
Hence, the series diverges, which implies:
$$\forall k\lim_{n\to \infty}\frac{n!}{n^k} = \infty$$
