Question: I want to find a natural number $N$ in terms of $m(\in\mathbb N)$, such that

$$n!>n^m \;, \forall n \ge N$$

Also, (how) can we prove that $n!-n^m$ is an increasing sequence for $n\ge N$?

I was just solving the problem that $n!>n^2$ for $n\ge 4$. I am trying to generalize the statement, but I don't know how to proceed.

Answer: 
Note the following:

1. $n\ge n$.
2. $(n-1)(n-2) \ge n$, for $n$ sufficiently large
3. $(n-3)(n-4) \ge n$, for $n$ sufficiently large
4. $(n-5)(n-6) \ge n$, for $n$ sufficiently large

If we do this for $m$ steps (and provided that $n$ is sufficiently large for all of them), we may multiply the LHS to get $n(n-1)(n-2)\ldots, (n-2m+2)$, which (provided $n>2m-2$) will be smaller than $n!$, while the RHS will be $n^m$. All of the inequalities are implied by the last one, which is $(n-2m+3)(n-2m+2)\ge n$. This rearranges to $n^2+(-4m+4)n+(2m-3)(2m-2)\ge 0$. Take the larger root of this quadratic, and $2m-2$ from above, and the larger of these will serve for $M$.
**Hint:**

You have this : $n! <n^n$

$n! <n^{n-1}$ , is this true$?$ Do you know that $n|(n-1)!$ if $n$ is composite$?$