Why is $\sum_{m>n}p(1-p)^{m-1}$ equal to $(1-p)^n \sum_{m>0}p(1-p)^{m-1}$?

Question

Suppose $p \in (0,1)$, $n,m \in \mathbb{N}$, we have:

$$\sum_{m>n}p(1-p)^{m-1} = (1-p)^n \sum_{m>0}p(1-p)^{m-1} $$

Why is this true? I can't seem to find the reasoning behind it.

Answer 1

$p \in (0,1)$ $ n,m,k \in \mathbb{N} $

Denote $m \mapsto k+n $

$$\sum_{m>n}p(1-p)^{m-1} = \sum_{k+n>n}p(1-p)^{k+n-1}= (1-p)^{n}\sum_{k>0}p(1-p)^{k-1}$$

Now changing back to index m
$k \mapsto m$

$$=(1-p)^n \sum_{m>0}p(1-p)^{m-1}$$
as was to be shown.
It's basically just changing and playing with indexes and bounds.