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?

Intuitively you know this must not converge because $n!$ requires $n$ multiplications whereas $n^k$ only involves $k$ multiplications.  Clearly as $n$ grows, the numerator grows more rapidly than the denominator.  The issue is that as $n$ grows both $n!$ gets bigger and $n^k$ gets bigger.  What we need to do is get a lower bound on $n!$.  All we need is $k$ terms in $n!$ that are bigger than $n$--but, of course, this isn't possible.
So instead of trying to get $k$ terms in $n!$ that are bigger than $n$, let's see what happens when you try to get $k$ terms that are bigger than a fraction of $n$.  Really any fraction will do, but let's make it concrete and try $\frac{n}{2}$:
Find $n$ such that $n - \frac{n}{2} \geq k \rightarrow n \geq 2k$.  Now if $n \geq 2k$ then we can say that $n! \geq \left(\frac{n}{2}\right)^k$.  This is saying that if $n$ is greater than $\frac{n}{2}$ by more than $k$ terms then $n!$ is greater than the minimum, $\frac{n}{2}$, multiplied by $k$ times.
This starts to form a possible solution.  We know that $n! \geq \left(\frac{n}{2}\right)^k$ therefore we can say that $\frac{n!}{n^k} \geq \left(\frac{1}{2}\right)^k$.  This certainly suggests that this sequence diverges but doesn't prove it.  The problem is in $\frac{n^k}{n^k}$.  Instead of $n \geq 2k$ we can go ahead and add anything to that and sum, for instance, $\frac{n}{2}\geq k + 1 \rightarrow n \geq 2k+2$.  In this case we get:
\begin{align}
n! \geq \left(\frac{n}{2}\right)^{2k+2} \\
\frac{n!}{n^k} \geq&\  \frac{\left(\frac{n}{2}\right)^{2k+2}}{n^k} \\
\geq&\ \frac{n^{k + 2}}{2^k}
\end{align}
Now clearly as $n$ goes towards infinity, this limit diverges.
This is a way to argue a squeeze theorem.  You could generalize this argument to be any fraction rather than $\frac{1}{2}$.
