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

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