Prove exponent(m)=e^{m} 

please show me how to do the third one, I just understand the 1st and 2nd, but i have no idea how to do the 3rd. thank you.
 A: Use induction as it is stated. Consider 
$$p(1): exp(1)= e^1$$
We know that
$$exp(1) = \sum_{n=0}^{\infty} \frac{1}{n!} = e=e^1$$
by the definition of $e$.
Now consider
$$p(k): exp(k) = e^k$$
$$\Rightarrow exp(k) \cdot e = e^{k+1}$$
$$\Rightarrow \sum_{n=0}^{\infty} \left[\frac{k^n}{n!}\right] \cdot e = e^{k+1}$$
$$\Rightarrow \sum_{n=0}^{\infty} \left[\frac{k^n}{n!}\right] \sum_{n=0}^{\infty} \left[\frac{1}{n!}\right] = e^{k+1}$$
$$\Rightarrow \sum_{n=0}^{\infty} \left[\sum_{i=0}^{\infty} \frac{k^i}{(n-i)!i!} \right] = e^{k+1} \mbox{(Cauchy product)}$$
$$\sum_{n=0}^{\infty} \left[\sum_{i=0}^{\infty} \frac{k^i}{(n-i)!i!} \right] = \sum_{n=0}^{\infty} \left[\sum_{i=0}^{\infty} \frac{k^i}{n!} {n \choose i} \right] $$
$$= \sum_{n=0}^{\infty} \left[ \frac{1}{n!} \left[\sum_{i=0}^{\infty} k^i {n \choose i} \right] \right] $$
$$= \sum_{n=0}^{\infty} \frac{(k+1)^n}{n!} \mbox{(binomial theorem)}$$
$$= e^{k+1}$$
$\therefore p(k) \Rightarrow p(k+1)$ and $exp(m) = e^m \ \forall m \in \mathbb{N}$ by induction. Now we need to use the corollary to prove for negative integers. This requires induction again. Consider
$$p(-1): exp(-1) = e^{-1}$$
$$\iff exp(-1) \cdot e = 1$$
We know that
$$exp(-1) \cdot e = \sum_{n=0}^{\infty} \left[ \frac{(-1)^n}{n!}\right] \sum_{n=0}^{\infty} \left[ \frac{1}{n!}\right]$$
$$= \sum_{n=0}^{\infty} \left[\sum_{i=0}^{\infty} \frac{(-1)^i}{(n-i)!i!} \right] \mbox{(Cauchy product)}$$
$$= \sum_{n=0}^{\infty} \left[ \frac{1}{n!} \left[\sum_{i=0}^{\infty} (-1)^i {n \choose i} \right] \right] $$
$$= \sum_{n=0}^{\infty}  \frac{0^n}{n!}  \mbox{(binomial theorem)}$$
$$= 1 + \sum_{n=1}^{\infty}  \frac{0^n}{n!}  = 1+0 = 1$$
Thus, we have proven $p(-1)$. Now it remains to do the inductive step. Consider
$$p(k): exp(k) = e^k $$ for all negative integers, k.
$$\Rightarrow exp(k) \cdot e^{-1} = e^{k-1}$$
$$exp(k) \cdot e^{-1} = \sum_{n=0}^{\infty} \left[\frac{k^n}{n!}\right] \cdot e^{-1}$$
$$= \sum_{n=0}^{\infty} \left[\frac{k^n}{n!}\right] \sum_{n=0}^{\infty} \left[\frac{(-1)^n}{n!}\right] $$
$$= \sum_{n=0}^{\infty} \left[\sum_{i=0}^{\infty} \frac{k^i(-1)^{n-i}}{(n-i)!i!} \right] \mbox{(Cauchy product)}$$
$$= \sum_{n=0}^{\infty} \left[ \frac{1}{n!} \left[\sum_{i=0}^{\infty} k^i(-1)^{n-i} {n \choose i} \right] \right] $$
$$= \sum_{n=0}^{\infty}  \frac{(k-1)^n}{n!} \mbox{(binomial theorem)}$$
$$=e^{k-1}$$
$\therefore p(k) \Rightarrow p(k+1)$ and $exp(m) = e^m \ \forall m \in \mathbb{Z} \backslash \{0\}$ by induction.
$$exp(0) = \sum_{n=0}^{\infty}  \frac{0^n}{n!}$$
$$= 1 + \sum_{n=1}^{\infty}  \frac{0^n}{n!} = 1+0 = 1 = e^0$$
So we conclude that $exp(m) = e^m \ \forall m \in \mathbb{Z}$.
A: Theorem 11.6. is probably
$$
\exp(x+y)=\exp(x)\exp(y)
$$
So for $n>1$ (induction starts with $n=1$)
$$
\exp(n+1)=\exp(n+1)=\exp(n)\exp(1)=e^ne=e^{n+1}
$$
if $n<0$ then $-n>0$
$$
1=\exp(0)=\exp((-n)+n)=\exp(-n)\exp(n)=e^{-n}\exp(n)
$$
