Proving equivalence between a summation expression and a power expression

Question

I have that $$2^{-(a+1)!}(1+\frac{1}{2^{1}}+\frac{1}{2^{2}}+...)=2(2^{-a!})^{a+1}\,\,\,\,\,\,\,\,(1)$$ Which I am trying to show is $\geq \sum_{b=a+1}^{\infty}2^{-b!}$ in order to prove $\sum_{b=0}^{\infty}2^{-b!}$ is transcendental by Liouville's theorem. However I am struggling to understand how we get equality in $(1)$.

Answer 1

$$2^{-(a+1)!}(1+\frac{1}{2^{1}}+\frac{1}{2^{2}}+...)=2^{-(a + 1)!} \left ( \sum_{i = 0}^\infty 0.5^i \right ) $$

$$ = 2^{-(a + 1)!} 2$$

$$ = 2^{-(a + 1) a!} 2$$

$$ = \left ( 2^{-a!} \right )^{a+1} 2 $$