I need to show that if $$ a_{n}=\sum_{i=0}^{n-1}\binom{n}{i}a_{i}, $$ then $$ a_{n}=\frac{1}{2}\sum_{k=0}^{\infty}\frac{k^{n}}{2^{k}}, $$
Solution says:
Since $$ 2a_{n}=\sum_{i=0}^{n}\binom{n}{i}a_{i} $$
then $$ A_{e}(x)=\frac{1}{2-e^{x}} = \frac{1}{2}\sum_{k=0}^{\infty}\left(\frac{e^{x}}{2}\right)^{k} = \frac{1}{2}\sum_{k=0}^{\infty}\sum_{n=0}^{\infty} \frac{k^{n}}{2^{k}} \cdot \frac{x^n}{n!} $$
How did we obtain the generating function $A_{e}(x)=\frac{1}{2-e^{x}}$? I understand all the other steps but can solve only simpler recurrences, without sums. Thanks.