From $32\cdot (\frac{1-.5^n}{1-.5})$ to $\frac{64(2^n-1)}{2^n}$ I have to prove that  $32\cdot (\frac{1-.5^n}{1-.5})$ is equal to $\frac{64(2^n-1)}{2^n}$
I know that they are equal because when I put them in my calculator the graphs are the same. However, I'd like to do this manually.
So far I have done:
$$32\cdot (\frac{1-.5^n}{1-.5}) = 32 \cdot (\frac{.5}{.5} + \frac{.5^n}{.5}) = 32 \cdot (\frac{.5\cdot.5^n}{.5}) = \frac{32\cdot.5\cdot.5^n}{.5}= \frac{8^n}{.5} $$
How do I solve this?

Don't be afraid of writing $.5 = \frac 12$
$32* (\frac{1-.5^n}{1-.5})=$
$2^5(\frac{1 - \frac 12^n}{\frac 12})=$
$2^6(1 - \frac 12^n)=$
$2^6(1 - \frac 12^n)\frac{2^n}{2^n}=$
$2^6\frac{2^n - 1}{2^n}=$
$64\frac{2^n - 1}{2^n}=$
