How can i make the following change to this infinite series?

Question

$$ e^z - 1 = \sum_{n=1}^\infty \frac {z^n}{n!} $$

Given the above function and its corresponding series expansion, is there anything i could do to the left side of the equation so that the infinite series looks like this instead???

$$ \sum_{n=1}^\infty (\frac {z^n}{n!})^{a} $$

that is to the power of $\mathbf A$ which would be any constant.

$$ (e^z-1)^a= \sum_{n=1}^\infty (\frac {z^n}{n!})^{a} $$

Would it be just like this? ^

Thank you very much for your time and help.

Answer 1

If you put a series to a certain power, it doesn't mean it's equal to the series of each term to that power.

I don't think there is a general expression for this series except when $a$ is 0 or 1.

Answer 2

For $a=2$, $$\sum_{n=0}^\infty \left(\frac{z^n}{n!}\right)^2 = I_0(2z)$$ where $I_0$ is a modified Bessel function. In general for positive integer $a$ you can write the series as a generalized hypergeometric function $$ {\mbox{$_0$F$_{a-1}$}(\ ;\,1,\ldots,1;\,{z}^{a})}$$