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: 
Since you are calling **for intuition, a term by term version**. Intuitively, the inverse factorial coefficient $\frac{1}{n!}$ is the most natural, as it yields "**some kind of invariance to differentiation**", since:

$$\left(\frac{x^n}{n!}\right)' = \left(n\frac{x^{n-1}}{n!}\right) = \frac{x^{n-1}}{(n-1)!} = \frac{x^{m}}{m!}$$
with $m= n-1$.

You may know that, and start with, a fundamental property of the natural exponential: it is equal to its derivative. 
So, suppose that  

$$ f(x) = \sum\_{n=0}^\infty a\_n{x^n}\,,$$
is equal to its derivative. Then formally (I am skipping issues on convergence), 
$$ f'(x) = \sum\_{n=1}^\infty n a\_n{x^{n-1}}\,,$$
thus by reindexing:
$$ f'(x) = \sum\_{n=0}^\infty (n+1) a\_{n+1}{x^{n}}\,,$$
then, one should have, term by term:
$$ a\_n= (n+1) a\_{n+1}\,,$$
hence
$$ a\_{n+1} = \frac{a\_n}{n+1} =\frac{a\_0}{(n+1)!}\,.$$
Since the exponential is the reciprocal to the $\log$, you require that $f(0)=1$, hence $a\_0=1$. So naturally, 

$$ f(x) = \sum\_{n=0}^\infty \frac{x^n}{n!}=e^x\,,$$
and 
$$ f(1) = \sum\_{n=0}^\infty \frac{1}{n!}=e\,.$$
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 $\mathbb{e}$