Exponential and Natural Log Power Series

Question

I am asked to prove $$e^{Y\log(1+Z)} = (1+Z)^Y$$ using power series definitions for both the exponential function and natural log. I am really stumped on this. Our hunt was using the nth derivative of the power series $e^{Y\log(1+Z)}$ and determine the coefficient on $Z^n$ but I am really stumped. Please help

Answer 1

$e^z=1+z+\frac{z^2}{2!}+...$ therefore $\frac{de^Z}{dz}=1+z+\frac{z^2}{2!}+...=e^z$.

$\log (1+z)=z-\frac{z^2}{2}+\frac{z^3}{3}-... $ therefore $\frac{d \log(1+z)}{dz}=1-z+z^2-...=\frac{1}{1+z}$.

Then $$\frac{d}{dZ}(e^{Y\log(1+Z)})=(e^{Y\log(1+Z)})\frac{Y}{1+Z}$$ and $$\frac{d}{dZ}(1+Z)^Y=(1+Z)^Y\frac{Y}{1+Z}.$$

By the quotient rule we then see that $\frac{e^{Y\log(1+Z)}}{(1+Z)^Y}$ has zero derivative and is therefore constant. Substitution of $Z=0$ shows that this constant is $1$ and therefore $$e^{Y\log(1+Z)} = (1+Z)^Y.$$