In my homework assignment it says: $\sum^\infty\_{i=1}p^2(1-p)^{2(i-1)}=\frac{p^2}{1-(1-p)^2}=\frac{p}{2-p}$

I don't see how $\sum^\infty\_{i=1}p^2(1-p)^{2(i-1)}=\frac{p^2}{1-(1-p)^2}$

I tried showing it, knowing that:

$\sum^\infty\_{i=1}p^2(1-p)^{2(i-1)} = \frac{p^2}{(1-p)^2}\sum^\infty\_{i=1}(1-p)^{2i}=$

$\frac{p^2}{(1-p)^2}\sum^\infty\_{i=0}(1-p)^{2i}-(1-p) = \frac{p^2}{(1-p)^2}\sum^\infty\_{i=0}((1-p)^{2})^{i}-(1-p) = \frac{p^2}{(1-p)^2} \cdot ( \frac{1}{(1-p)^2} -1 )$ 

it does not seem to be $\frac{p^2}{1-(1-p)^2}$

is there a mistake I've made along the way? Can someone help?