Prove $\frac{1}{\exp(z)}=\exp(-z)$ by Taylor series Prove $$\frac{1}{\exp(z)}=\exp(-z)$$ by Taylor series.
Since $$\exp(z)=\sum_{k \geq 0}\frac{z^k}{k!}$$
What can we do with $\frac{1}{\exp(z)}$?
 A: Theorem
$$e^{ax}.e^{bx}=e^{(a+b)x} $$
(We will use the series definition and the cauchy product of two series, it works on formal power series also)
Proof:
We will take the product of the two series
$e^{ax}=\sum\limits^{\infty}_{k=0}\frac{a^{k}}{k!}x^{k}$  and
$e^{bx}=\sum\limits^{\infty}_{k=0}\frac{b^{k}}{k!}x^{k}$
$$\sum\limits^{\infty}_{k=0}\frac{a^{k}}{k!}x^{k} \sum\limits^{\infty}_{k=0}\frac{b^{k}}{k!}x^{k} =\sum\limits^{\infty}_{k=0}c_{k}x^{k} $$
where $c_{k}=\sum\limits^{k}_{p=0}\frac{a^{p}}{p!}
\frac{b^{k-p}}{(k-p)!}$ multiplying by $k!$ we have
$c_{k}k!=\sum\limits^{k}_{p=0}k!\frac{a^{p}}{p!}
\frac{b^{k-p}}{(k-p)!}=\sum\limits^{k}_{p=0} {k \choose
p}a^{p}b^{k-p}$ by the binomial theorem $c_{k}k!=(a+b)^{k}$ so
$c_{k}=\frac{(a+b)^{k}}{k!}$, then we have
$$\sum\limits^{\infty}_{k=0}\frac{a^{k}}{k!}x^{k} \sum\limits^{\infty}_{k=0}\frac{b^{k}}{k!}x^{k} =\sum\limits^{\infty}_{k=0}\frac{(a+b)^{k}}{k!}x^{k} $$
Corollary
Take $a=1$, $b=-1$, then the result is $e^{0}=1$ by the series. Then $e^{-x}=\frac{1}{e^x}.$
A: So you will need to make use of discrete convolution, i.e. a way to take the product of two series and end up with a series with two indices (think of it as an extension of polynomial multiplication, or in parallel with its continuous counter part, integral convolution). 
The formula for convolution gives
$$
\exp(z)\exp(-z)=\left(\sum_{i=0}^\infty\frac{z^k}{k!}\right)\left(\sum_{i=0}^\infty\frac{(-z)^j}{j!}\right)\\
=\sum_{k=0}^\infty\sum_{l=0}^k\frac{z^l(-z)^{k-l}}{l!(k-l)!}
$$
now notice that the summand looks a lot like the binomial expansion of 
$$
(z-z)^k=0^k
$$
but off by a $k!$. So we divide by $k!$ to compensate and find that the product is equal to 
$$
\sum_{k=0}^\infty\frac{0^k}{k!}=1
$$ 
$e^0=1$ if you like, or since $0^k=0$ for every $k>1$ but is $1$ for $k=0$.
