For my course in complex analysis we have to prove that the trivial relation $e^{\ln{z}}=z$. We are given the series for $\ln z$:
$$f(w)=\sum_{n=0}^\infty (-1)^{n+1}\frac{w^n}{n}$$
$$\ln z = f(z-1)$$
I know that the series for $e^x$ is
$$e^{x}=\sum_{n=0}^\infty \frac{x^n}{n!}$$
I tried to solve
$$e^{\ln{z}}=\sum_{n=0}^\infty \frac{\left(\sum_{m=0}^\infty (-1)^{m+1}\frac{(z-1)^m}{m}\right)^n}{n!}$$
But just inserting the previous series into this does not yield a very convenient result. I think that if we expand the first power series around $z=0$ we would have already a problem (this makes sense since also $\ln 0$ does not exist. How and with what technique is this problem solved?
Edit (clarification):
We need to prove the relation given using the power series for $\ln{z}$ as definition.