1. Sep 12, 2011

### lackrange

The actual question is prove that $$|\alpha|!\le n^{|\alpha|}\alpha!$$ where
$$\alpha=(\alpha_1,.....\alpha_n)$$ is a multi-index (all non-negative) and $$|\alpha|=\alpha_1+\cdots +\alpha_n$$ and $$\alpha!=\alpha_1!\cdots \alpha_n!$$ so I am trying to do it by induction on the number of elements $$n$$ in $$\alpha$$...so I am trying to prove that $$(a+b)!<2^{a+b}a!b!$$ I have tried to do this by induction on the value of b (the inequality is obvious for b=0 or 1), and other ways, but nothing is working (been trying for close to a week).

(ps. how do I make it so that after I write in latex it doesn't skip a line like that?)

2. Sep 12, 2011

### micromass

Are you allowed to use Stirling approximation??

By the way, use [itex ] if you don't want newlines.

3. Sep 12, 2011

### micromass

Anyway, if you're not allowed to use Stirling approximation, just notice that your inequality is equivalent to

$$\binom{a+b}{a}<2^{a+b}$$

Now you can use a combinatorial argument.

4. Sep 12, 2011

### lackrange

I believe I am allowed to use stirling's approximation, can you suggest a way? (it's only approximate for large n).

Anyway, I will try the other way in the mean time, thanks.

5. Sep 12, 2011

### lackrange

Never mind, I got it! You were a huge help, thank you!