Calculating Bernoulli Numbers from $\sum\limits_{n=0}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$ How is the Bernoulli numbers? For example, found that in internet 
$$\sum_{n=0}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$$
but if I want to find $B_2$ then
$$B_0+B_1x+\frac{B_2x^2}{2}+\sum_{n=3}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$$
and I believe this is not much help. 
I want to learn how to calculate Bernoulli numbers to learn how to calculate $\zeta(2n)$.

Note that
$$\frac{e^z-1}z=\frac1 z\sum_{n=1}^\infty\frac1{n!}z^n=\sum_{n=1}^\infty\frac1{n!}z^{n-1}=\sum_{n=0}^\infty\frac1{(n+1)!}z^n$$
and we can use Mertens’ multiplication theorem to get
$$1=\left(\sum_{n=0}^\infty\frac{B_n}{n!}z^n\right)\left(\sum_{n=0}^\infty\frac1{(n+1)!}z^n\right)=\sum_{n=0}^\infty\sum_{k=0}^n\left(\frac{B_k}{k!}\frac{1}{(n-k+1)!}\right)z^n$$
By the identity theorem, the $n=0$ term on the right must equal $1$ while all other
terms must vanish. The $n=0$ term on the right is just $B_0$, so $B_0 = 1$, and for
$n > 1$, we must have $\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}=0$. Multiplying this by $(n + 1)!$ we get
$$0=\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}=\sum_{k=0}^n\frac{(n+1)!}{k!(n-k+1)!}B_k=\sum_{k=0}^n\binom{n+1}kB_k$$
and adding $B_{n+1}=\binom{n+1}{n+1}B_{n+1}$ to both sides of this equation, we get
$$B_{n+1}=\sum_{k=0}^{n+1}\binom{n+1}kB_k$$
The right-hand side might look familiar from the binomial formula. Recall from
the binomial formula that for any complex number $a$, we have
$$(a+1)^{n+1}=\sum_{k=0}^{n+1}\binom{n+1}ka^k1^{n+1-k}=\sum_{k=0}^{n+1}\binom{n+1}ka^k$$
Notice that the right-hand side of this expression is exactly the right-hand side of
the previous equation if put $a = B$ and we make the superscript $k$ into a subscript
$k$. Thus, if we use the notation $\Doteq$ to mean equals after making superscripts into
subscripts, then we can write
$$\boxed{B^{n+1}\Doteq (B+1)^{n+1},n=1,2,3,...,B_0=1}$$
Use recent identity, one can in principle find all the Bernoulli numbers: When
$n = 1$, we see that
$$B^2\Doteq(B+1)^2=B^2+2B^1+1\Rightarrow0=2B_1+1\Rightarrow B_1=\frac{-1}2.$$
When $n = 2$, we see that
$$B^3\Doteq(B+1)^3=B^3+3B^2+3B^1+1\Rightarrow3B_2+3B_1+1=0\Rightarrow B_2=\frac1 6.$$
