Here, at point 5, you can find a proof that, for a variable with a geometric distribution,
$P(N>n) = (1-p)^n$
The proof is a follow:
$P(N>n) = \sum_{k=n+1}^{+\infty}P(Y=a)=\sum_{k=n+1}^{+\infty}(1-p)^{k-1}p= \frac{p(1-p)^n}{1-(1-p)}=(1-p)^n $
I am unclear about why the following step is possible:
$\sum_{k=n+1}^{+\infty}(1-p)^{k-1}p= \frac{p(1-p)^n}{1-(1-p)}$
I am aware that it exploits the following formula for geometric series: $\sum_{i=1}^{+\infty}r^i = \frac{r}{1-r}$
However, to use that formula, you need and index starting at 1 and that index has to be alone in the exponent. In this situation, it is possible to perform an "index shift" to achieve to the desired form. However, I don't see what index shift was performed. Moreover, the result is actually of the form $\frac{r^n}{1-r}$, which is slightly different, probably due to the index shift that was done. Any insight would be appreciated!
Firt thid id the idea behind all this: $$\begin{align}\sum_{k=n+1}^{+\infty}(1-p)^{k-1}p&=p(1-p)^n\sum_{k=\color{#FF0000}{n+1}}^{+\infty}(1-p)^{\color{#FF0000}{k-n-1}}\tag{1}\\&=p(1-p)^n\sum_{\color{#0000FF}{i=0}}^{+\infty}(1-p)^{\color{#0000FF}i} \tag{2}\\&= \frac{p(1-p)^n}{1-(1-p)}\tag{3}\end{align}$$
we change the index $\color{#0000FF}{i=k-n-1}$, and in the last transition we used effectively (for $r=1-p$):$$\sum_{i=0}^{+\infty}r^i=\frac{1}{1-r} \tag{4}$$
Further explanations
You don't have to worry if we start from $1$ or $2$ in fact if we multiply the equation $4$ by $r$ we obtain: $$\frac{r}{1-r}=\sum_{\color{#FF0000}{i=0}}^{+\infty}r^{\color{#FF0000}{i+1}}=\sum_{\color{#0000FF}{k=1}}^{+\infty}r^{\color{#0000FF}{k}} \tag{5}$$
And we can modify this slightly to obtain: $$\frac{r^n}{1-r}=\sum_{\color{#FF0000}{i=0}}^{+\infty}r^{\color{#FF0000}{i+n}}=\sum_{\color{#0000FF}{k=n}}^{+\infty}r^{\color{#0000FF}{k}} \tag{6}$$