I'm self teaching Complex Analysis, and am working on this question:
Show that the series expansion of $\exp(z)$ has radius of convergence 1. Let $f(z)$ be the function to which the series $\sum\limits_{n=1}^{\infty} \frac{z^n}{n}$ converges on the domain $$D = \{z \in \Bbb C : |z| < 1\}$$ Show that $\exp(f(z)) = \frac{1}{1-z}$
I've done the radius of convergence part by applying the ratio test, however, am not sure how to proceed next. Would be grateful for some guidance, thanks.
Continued comment.
We could try doing this without knowledge of the function $\log$. Can we combinatorially combine the power series to get this conclusion? That is, show $$ \sum_{m=0}^\infty \frac{1}{m!}\left(\sum\limits_{n=1}^{\infty} \frac{z^n}{n}\right)^m = \sum_{k=0}^\infty z^k $$An exercise in "generatingfunctionology".
Equivalent statement: Fix a nonnegative integer $k$. Consider all finite sequences $(n_1,n_2,\dots,n_m)$ of positive integers with sum $k$. Then $$ \sum \frac{1}{m!n_1\cdots n_m} = 1 $$ summed over all those finite sequences.
For example, if $k=3$, then the possibilities are: $3, 2+1, 1+2, 1+1+1$, and $$ \frac{1}{1! \cdot 3}+\frac{1}{2!\cdot 2\cdot 1} +\frac{1}{2!\cdot 1\cdot 2}+ \frac{1}{3!\cdot1\cdot1\cdot1} = \frac{1}{3}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6} = 1 $$ Can you do this for all $k$? \begin{align} k=0:&\qquad \frac{1}{0!} = 1 \\ k=1:&\qquad \frac{1}{1!\cdot 1} = 1 \\ k=2:&\qquad \frac{1}{1!\cdot 2} + \frac{1}{2!\cdot1\cdot1} = 1 \end{align}