# Comparison between factorials with power

I try to understand some logic proof and I stumbled upon following:

$$\frac{(N+2m)!}{N!\ m!\ m!} < \frac{({N+2m})^{2m}}{m!\ m!}$$

I don't know why it's true.

This is from a book called "A First Course in Logic" by S.Hedman. I think that it will be really obvious, when I see it, but I wasn't able to work that out.

First, $$n!=n \times (n-1) \times (n-2) \times \cdots \times 1< n \times n \times n \times \cdots \times n = n^n$$
Hence $$\frac{\left(N+2m\right)!}{N!}<\left(N+2m\right)^{2m+N-N}=\left(N+2m\right)^{2m}$$ Noticing that in fact : $$\frac{\left(N+2m\right)!}{N!}=\left(N+2m\right)\left(N+2m-1\right)\dots\left(N+1\right)$$
You have: \begin{aligned} (N+2m)! &= 1 \times 2 \times 3 \times \dots \times N \times (N+1) \times (N+2) \times \dots (N+2m)\\ &= N! \times (N+1) \times (N+2) \times \dots \times (N+2m)\\ &< N! \times (N+2m)^{2m} \end{aligned}
Because for $1 < k \le 2m$, $N+k < N+2m$.