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 one approach.

Let $b > 1$. If you compute the derivative of the function $b^x$, you find that the answer is just $b^x$ multiplied by an (annoying) constant.

There is a value of $b$ for which this constant is equal to $1$. That's nice! With this special value of $b$, the derivative of $b^x$ is just $b^x$, the same thing we started with. That's a very neat property for a function to have.

This special value of $b$ is $e = 2.718 \ldots$.

It is now easy to compute the Taylor series of the function $e^x$ (centered at $0$). We find that
\begin{equation}
\tag{$\spadesuit$} e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.
\end{equation}
This comes directly from the Taylor series formula
\begin{equation}
f(x) = f(x\_0) + f'(x\_0)(x - x\_0) + \frac{f''(x\_0)}{2!}(x - x\_0)^2 + \cdots.
\end{equation}

Plugging $x = 1$ into ($\spadesuit$) yields
\begin{equation}
e = \sum\_{n=0}^{\infty} \frac{1}{n!}.
\end{equation}
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
$$\frac{1}{e^x}=1-\frac{x^1}{1!}+\frac{x^2}{2!}-...$$

---

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:
$$f(n)=1+\frac{ln(f(n))^1}{1!}+\frac{ln(f(n))^2}{2!}+..$$