Question: A finite Hausdorff topological group, has discrete topology and every discrete group is totally disconnected. I look for an example of a non-Abelian, finite, connected non-Hausdorff group . I think every non-Abelian, finite, connected groups must be discrete.

Answer: 
Let $x \in \mathbb{R}^+$.

$f:x\longrightarrow \exp(x) \in C^\infty(\mathbb{R}, \mathbb{R})$, hence we can write $\forall n \in \mathbb{N}$ :

$$\left|f(x)-\sum\_{k=0}^n f^{(k)}(0)\frac{x^k}{k!}\right| \leq \frac{|x|^{n+1}}{(n+1)!}I\_{n+1} \text{ where } I\_{n+1}=sup\_{[0,x]}|f^{n+1}|$$

And $\forall t \in [O,x] f^{(n)}(t)=\exp(t)$ which means $I\_{n+1}=\exp(x)$

But $\displaystyle\lim \_{n\to \infty }\left(\frac{|x|^{n+1}}{(n+1)!}\exp(x)\right)=\:0$

So $\displaystyle\lim \_{n\to \infty }\left(\sum\_{k=0}^n\frac{x^{k}}{k!}\right)=\exp(x)$

Just take $x = 1$ for $\mathbb{e}$
By definition   

$$\frac{d}{dx}e^x = e^x$$
therefore
$$\int\_0^xe^t\*1dt=e^x-1$$
we can use integration by parts which state that:
$$\int{f(x)g(x)dx}=f(x)\int g(x) - f^`(x)\int\int g(x)+...$$
let f(x)= e^x, g(x) = 1
$$e^x\int1-e^x\int\int1+...=e^x-1$$
we simplify to get:
$$\frac{1}{e^x}=1-\int1+\int\int1-\int\int\int1-...$$
we know that $\int\int..n..\int1dx=\frac{x^n}{n!}$ (basic integration rule)   

therefore
$$\frac{1}{e^x}=1-\frac{x^1}{1!}+\frac{x^2}{2!}-...$$

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let substitute x = -n, we know that (-x)^n = -x^n when n is odd
$$e^n = 1+\frac{x^1}{1!}+\frac{x^2}{2!}+..$$
in fact, if we are to substitute x = ln(f(n)), n > 0 we get a nice series:
$$f(n)=1+\frac{ln(f(n))^1}{1!}+\frac{ln(f(n))^2}{2!}+..$$