Prove that $n! > n^5$ I'm trying to prove that $n! &gt; n^5$ for large enough values of n. While it seems obvious that this should be true, I have no idea how to prove it rigorously. 

EDIT:

So, looking at the comments, here's what I'm thinking now. What do folks think? Is the following reasonable?

The idea is to take the limit of the ratio of $(n+1)!/n!$ vs $(n+1)^5/n^5$ to show that the former is much larger than the latter, and thus that when n &gt; M, where M is some positive integer, $n!$ must be growing much larger than $n^5$:

Let us assume $\frac{(n+1)!}{n!} &gt; \frac{(n+1)^5}{n^5}$

$$
\lim_{n\to\inf} \frac{(n+1)!}{n!} &gt; \frac{(n+1)^5}{n^5}\\
= \lim_{n\to\inf} n+1 &gt; (\frac{n+1}{n})^5\\
= \lim_{n\to\inf} n+1 &gt; (1 + \frac{1}{n})^5\\
\ldots
=\lim_{n\to\inf} n+1 &gt; (1 + 3/n + 8/n^2 + 10/n^3 + 5/n^4 + 1/n^5)\\
= inf &gt; 1
$$

(Edited [again]: I did not know you could use latex)
For $n \geq 10$, we have
$$n! = n(n-1) \cdots (n-4)(n - 5)! \geq \left(\frac{n}{2}\right)^5 (n - 5)! = n^5\left(\frac{(n -5)!}{32}\right) &gt; n^5.$$
(The bound above is not sharp, but it's sufficient for the problem. Also, the same argument shows that for any $p &gt; 0$, we have $n! &gt; n^p$ for sufficiently large $n$.)
The first derivative of those should be useful in your proof.  Start with n = $10$ for example.

Someone asked me to explain further but I am not a math person so it will be a simple explanation.

For n = $10$ we have $10$! vs. $10^5$ which is $3,628,800$ vs. $100,000$ so it is already larger at the point.  If we increase n to $11$, the $10$! increases by a factor of $11$ but the $10^5$ increases by $1.1^5$ which is only about $1.61$.  If we increase n to $12$ you can see a "breakaway" between them so just prove this breakaway by showing the "discrete 1st derivative" of n! is increasing greater than that of $n^5$ and you got it.