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 $M$.
You can get rid of the factorial by taking logs:
$$ \log(n!) = \sum\_{j=1}^n \log(j).$$
Then you can estimate this sum using the integral since $\log$ is monotone:
$$\sum\_{j=1}^n \log(j) \ge \int\_1^n \log(x) dx = n\log(n) -n +1.$$

Then 
$$ n! \ge n^m \Leftrightarrow \log(n!) \ge m \log(n),$$
and to prove this it is sufficient to prove that 
$$ n\log(n) -n +1 \ge m \log(n) \Leftrightarrow (n-m)\log(n) \ge n-1.$$
Assume $n \ge m+1$. Then the last inequality reduces to 
$$ \log(n) \ge \frac{n-1}{n-m}.$$
But 
$$ \max\_{n\ge m+1} \frac{n-1}{n-m} = m,$$ 
so the previous inequality will hold if 
$$ \log(n) \ge m \Leftrightarrow n \ge e^m.$$
Note that $n \ge e^m$ implies $n \ge m+1$ since $e^x \ge x+1$ for $x \ge 0$. So, a *sufficient* condition for $n! \ge n^m$ is $n \ge e^m.$

This is not so great, so instead we can try restricting to, say $n \ge 2m$. Then
$$ \max\_{n\ge 2m} \frac{n-1}{n-m} = \frac{m-1}{m},$$ so a sufficient condition is then
$$ \log(n) \ge \frac{m-1}{m} \Leftrightarrow n \ge \exp\left( \frac{m-1}{m}\right).$$ So, a *better sufficient* condition is
$$ n \ge \max\left\{2m, \exp\left( \frac{m-1}{m}\right) \right\}.$$ This is more in line with the other solutions since $\frac{m}{m-1}$ remains bounded as $m \to \infty$.