Why $\bigg\lfloor \frac{n^m}{\binom{n}{m}}\bigg\rfloor=m!$ for $n>(m+1)^{m+2}$? Let $n,m\in \mathbb{N}_0$. I need to prove that if $n>(m+1)^{m+2}$ then
$$ \Bigg\lfloor \frac{n^m}{\binom{n}{m}}\Bigg\rfloor=m!.$$
Since
$$ \frac{n^m}{\binom{n}{m}}=\frac{n^m(n-m)!}{n!}m!=m!+\bigg(\frac{n^m(n-m)!}{n!}-1\bigg)m!,$$
I have tried to see when
$$\frac{n^m(n-m)!}{n!}-1<\frac{1}{m!}.$$
Can anyone help me or give me a hint? Thanks in advance!
 A: I think your approach is fine, so I've continued with it below.
We can rewrite the fraction $\frac{(n-m)!}{n!}$ as follows:
$$\frac{(n-m)!}{n!} = \frac{1}{n(n-1)(n-2)...(n-m+1)} = \prod_{k = 1}^m{\frac{1}{n-k+1}}$$
And $n^m$ can be written as
$$\prod_{k = 1}^m{n}$$
Then combining the products gives
$$\frac{n^m(n-m)!}{n!} = \prod_{k = 1}^m{\frac{n}{n-k+1}} = \prod_{k = 1}^m{\left(1 + \frac{k - 1}{n-k+1}\right)}$$
which can be expanded out as
$$1 \cdot \left(1 + \frac{1}{n-1}\right) \cdot \left(1 + \frac{2}{n-2}\right) \cdot  ... \left(1 + \frac{m-1}{n-m+1}\right)$$
Subtracting the 1 as you've done, leaves us with an expansion of $2^{m-1}$ terms (since each factor multiplies the number of terms by 2).
By inspection, we see that the largest possible size of a single term is $\frac{m-1}{n-m+1}$
Therefore, the remainder $\left(\frac{n^m(n-m)!}{n!} - 1\right)$ can be bounded by
$$2^m \cdot \frac{m}{n-m}$$
We want this to be less than $\frac{1}{m!}$, thus
$$2^m \cdot \frac{m}{n-m} < \frac{1}{m!}$$
From here, its quite easy to rearrange for n. Then taking logs of both sides, and applying Stirlings Formula gives you the bound
$$n > m^m$$
which is slightly better than the one in your question.
I hope this works for you, and was easy enough to follow :)
