Why is $ \sum_{k = 0}^{n}{n \choose k} (-p)^k(1-p)^{n-k} = (1 - 2p)^n $

Question

As the title suggests, I am trying to see why this identity holds.

$$ \sum_{k = 0}^{n}{n \choose k} (-p)^k(1-p)^{n-k} = (1 - 2p)^n $$

Answer 1

Write $(1-2p)$ as $(1-p-p)$ let $a=1-p$ and $b=-p$ and apply binomial theorem to $(a+b)^n$

Answer 2

binomial expansion of $$((-p)+(1-p))^n=(1-2p)^n $$