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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?

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    $\begingroup$ This is Newton's binomial formula $\endgroup$ Sep 19 '20 at 7:32
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Hint

You have that

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

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