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: 
By definition,
$$e=\lim\_{n\to\infty}\left(1+\frac1n\right)^n.$$
Using the binomial theorem, the $k^{th}$ term of the development is
$${\binom nk}\frac1{n^k}=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!.n.n.n\dots n},$$
and 
$$\lim\_{n\to\infty}{\binom nk}\frac1{n^k}=\frac1{k!}.$$

For example, 
$$\left(1+\frac1{1000}\right)^{1000}=\frac1{0!}+\frac1{1!}+\frac{0.999}{2!}+\frac{0.997002}{3!}+\frac{0.994010994}{4!}\dots$$
Let $$I\_n=\int\_1^e\frac{(\ln x)^n}{x^2}dx.$$ Using integration by parts we have $$I\_{n+1}=-\frac{1}{e}+(n+1)I\_n.$$
Using induction we can prove that
$$\frac{I\_n}{n!}=1-\frac{1}{e}\sum\_{k=0}^n \frac{1}{k!}$$
since $(I\_n)\_n$ is bounded, we let $n\to \infty$ we get
$$\sum\_{k=0}^\infty \frac{1}{k!}=e.$$