1. ## inverse

how does one prove that 1/(e^z) = e^(-z)?

it seems like such an obvious statement.

2. Well it's obvious if you consider it a definition. However if you've defined $e^x$ as the power series $e^x=1+x+\frac{x^2}{2!}+...$ then it's less obvious. It's possible to prove, using the power-series definition, that $e^xe^y = e^{x+y}$, using an essentially combinatorial argument (which I can supply if you wish). It also follows from the power-series definition that $e^0=1$. From these two identities it follows that

$e^ze^{-z}=e^{z-z}=e^0=1$

So that $e^{-z}$ is the multiplicative inverse of $e^z$, i.e. $e^{-z}=(e^z)^{-1}$. Note also that the above shows that $e^z$ is never 0, because it is always invertible.

Hope that helps!

3. Okay well, apparently my post was useful so I'll supply the omitted details.

We know that

$e^x=\sum_{j=0}^\infty\frac{x^j}{j!}$

$e^y=\sum_{k=0}^\infty\frac{y^j}{k!}$

so $e^xe^y=\left(\sum_{j=0}^\infty\frac{x^j}{j!}\right )\left(\sum_{k=0}^\infty\frac{y^j}{k!}\right) = \sum_{j,k}\frac{x^jy^k}{j!k!} = \sum_{n=0}^\infty\sum_{j+k=n}\frac{x^jy^k}{j!k!}$

$=\sum_{n=0}^\infty\frac{1}{n!}\sum_{j+k=n}\frac{n! }{j!k!}x^jy^k = \sum_{n=0}^\infty\frac{1}{n!}(x+y)^n = e^{x+y}$

by the Binomial theorem.