There is a explicit form that admit Bernoulli numbers rationals but there is another definition where the Bernoulli numbers are $B_n$, such that $\displaystyle \frac{x}{e^x-1}= \sum_{n=0}^\infty B_n \frac{x^n}{n!}$.
How can I prove that?? or it's equivalence?
On
We have $$\frac{x}{e^x-1} = \frac{x}{x + x^2/2 + x^3/6 + x^4/24 + \cdots} = \frac{1}{1 + x/2 + x^2/6 + x^3/24 + \cdots}.$$ Now, if we let $f(x) := \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \cdots$, then $$\frac{x}{e^x-1} = \frac{1}{1 + f(x)} = 1 - f(x) + (f(x))^2 - (f(x))^3 - \cdots.$$ In this expansion, each power of $f(x)$ has rational coefficients, and since $f(x)$ is divisible by $x$, the final coefficient of $x^n$ is the same as the coefficient of $x^n$ in $1 - f(x) + (f(x))^2 - (f(x))^3 + \cdots + (-1)^n (f(x))^n$.
From the definition you gave for Bernoulli numbers, $$x=\left(\sum\limits_{n=0}^{\infty} B_n\dfrac{x^n}{n!}\right)(e^x-1)=\left(\sum\limits_{n=0}^{\infty} B_n\dfrac{x^n}{n!}\right)\left(\left(\sum_{n=0}^{\infty} \dfrac{x^n}{n!}\right)-1\right)$$
$$=\sum\limits_{n=0}^{\infty}\left(\left(\sum\limits_{j=0}^n\dfrac{B_j}{j!(n-j)!}\right)-\dfrac{B_n}{n!}\right)x^n.$$
Equating the coefficients of $x$ shows that $B_0=1,$
and equating the coefficients of $x^{n+1}$ for $n>0$ shows that $$0=\left(\sum\limits_{j=0}^{n+1}\dfrac{B_j}{j!(n+1-j)!}\right)-\dfrac{B_{n+1}}{(n+1)!}$$
i.e., $$B_{n+1}=\sum\limits_{j=0}^{n+1} \binom{n+1}j{B_j}.$$
This recurrence relation can be used to calculate $B_n$ for $n>0$,
since $B_{n+1}$ can be cancelled from both sides.
In any event, it is clear from this that the Bernoulli numbers are rational.