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 $M$.
Try to use induction twice. Firstly for $m$, and then for $n$.
The plan will look like this:

1. Base case. We need to proove that $n!>n^2, \forall n>N$ and find $N$. It's quite easy.
2. Inductive step. Let's assume that we know $N$ for $m$. We need to find $N'$ in terms of $N$, such that $n!>n^{m+1},\forall n>N'$. Find $N'$ and proove that $(N')!>(N')^{m+1}$ (basis). Then proove that $(n+1)!>(n+1)^{m+1}$ holds if $n!>n^{m+1}$ is true (hint: you also know that $(n+1)!>(n+1)^m$ from first induction (it's true, because $n>N'$ and $N'>N$, hence $n>N$), so you can show that $(n+1)^m<n^{m+1}$). Finally, divide $(n+1)!>(n+1)^{m+1}$ by $n+1$. It's easy to show that $n!>(n+1)^m$, since $n^{m+1}>(n+1)^m$.