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: 
Here is a complete rigorous proof, not using power series/exponential function. Suppose we are just learning about limits and series - we didn't learn about uniform convergence, etc.

Take this as the definition of $e$: $e = \lim\_{n \to \infty} (1 + \frac 1 n)^n$. Then it is trivial to show that for any FIXED $N$, no matter how large (but FIXED), we also have $\lim\_{n\to\infty}(1 + \frac 1 n)^{n+N} = e$ as well.

Let $S\_n = \sum\_{k=0}^n\frac 1 {k!}$. By the binomial theorem, $(1+\frac 1 n)^n \le S\_n$ for all $n$, so we get that the sum of the infinite series is at least $e$ (comparison term by term: ${n \choose k} \cdot \frac 1 {n^k} \le \frac 1 {k!}$ for all $k$ from $0$ to $n$). This is the easy part.

For the other part, FIX a natural number N (however large, but FIXED). Again by the binomial theorem, and through the same kind of computation, we have $S\_N \le (1+\frac 1 n)^{n+N}$. We only need the first $N+1$ terms of the expanded sum from the binomial theorem; $\frac 1 {k!} \le { n+N \choose k} \cdot \frac 1 {n^k}$. Now keep $N$ fixed and let $n \to \infty$; this shows that $S\_N \le e$. Finally, since this is true for EVERY $N$, we get the opposite inequality: the sum of the series of inverses of factorials is $\le e$. By the way, this also shows the series converges (which we could see in other ways, but this by itself is a complete proof - we showed the partial sums are bounded by $e$).
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:
$$f(n)=1+\frac{ln(f(n))^1}{1!}+\frac{ln(f(n))^2}{2!}+..$$