# Help with $\sum_{w=0}^{n} {n \choose w}(q-1)^w=q^n$

This comes from coding theory by Freudenberger et al. p33

$$\sum_{w=0}^{n} {n \choose w}(q-1)^w=q^n$$

So my start: $$\sum_{w=0}^{n} {n \choose w}\sum_{x=0}^{w}{w\choose x}(q)^x(-1)^{w-x}$$ now the first term could cancel 1 term from the inner sum using $$\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}$$ but I'm not sure this helps with the all the other terms. What's the quick solution that I don't see?

• This is Newton's binomial formula Sep 19 '20 at 7:32

$$q^n=((q-1)+1)^n.$$