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: 
For the first question: Note that 

$k \cdot (n-k)>n \Leftrightarrow kn-n>k^2 \Leftrightarrow n> \frac{k^2}{k-1}=k+1+\frac{1}{k-1} $

Thus, for all $k \neq 2$ and $n> k+2$ we have $k \cdot (n-k)>n$.

Setting this for $k=2, k=3, ..., k=m$ we get that if $n>2(m+2)$ we have

$$n! =1 \cdot [2 \cdot (n-2)]\cdot [3 \cdot (n-3)]...\cdot [m \cdot (n-m)] .. \cdot n >n^m$$

For the second question, I don't think the sequence is increasing, it is just eventually increasing. To prove this, just observe that if $n> 2^m$ and $n>N$ then $ n! \cdot n > 2^mn^m$ thus

$$(n+1)!-n!=n! \cdot n > (2n)^m > (n+1)^m > (n+1)^m-n^m$$
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 $M$.