How do we get the final formula of the Bernoulli number? I was trying to understand Bernoulli numbers. When I googled, I found this link. 
It starts by saying that, The Bernoulli numbers are defined via the coefficients of the power series expansion of 
$\frac{t}{e^{t}-1}$,
then they write the expansion,for $m \geq 0$.
$\frac{t}{e^{t}-1} = \sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m}$ (1)
Then, the tutorial says that we multiply both sides by $(e^{t} -1)$ and get 
$B_0=1$, 
$B_m= -\frac{1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}B_k$. (2)
How is this happening? 
I tried doing this with hand and I get a recursive relation, that looks like, 
$\frac{t}{e^{t}-1} = \sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m}$
$\implies t = e^{t}\sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m} - \sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m}$
$\implies t = e^{t}\sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m} - \frac{t}{e^{t}-1}$
What next?
How do I get to (2)? 
I am new to number theory, please excuse me if the doubt is very basic. Any kind of help will be appreciated. :)

Using Maclaurin series for $e^t$ we get that
$$
e^t - 1 = \sum_{k=1}^{\infty}\frac{t^k}{k!}.
$$ 
Now we get 
$$
t = \left(\sum_{m=0}^{\infty}\frac{B_m}{m!}t^m\right)\left(\sum_{k=1}^{\infty}\frac{1}{k!}t^k\right)
$$
or, dividing by $t$
$$
1 = \left(\sum_{m=0}^{\infty}\frac{B_m}{m!}t^m\right)\left(\sum_{k=1}^{\infty}\frac{1}{k!}t^{k-1}\right) = \left(\sum_{m=0}^{\infty}\frac{B_m}{m!}t^m\right)\left(\sum_{k=0}^{\infty}\frac{1}{(k+1)!}t^{k}\right).
\tag1 $$
Let's see which coefficient $a_n$ has the $t^n$ in the expanded right-hand part of this equation for some $n > 0$ (we expect this coefficient to be a zero). $t^n$ may appear if $t^m$ from the first sum multiplies the $t^{n-m}$ from the second one for some $0\le m\le n$. Thus 
\begin{align}
a_n = \sum_{m=0}^{n}\frac{B_{n-k}}{m!(n-m+1)!}= \sum_{m=0}^{n} {n+1\choose m} \frac{B_{m}}{(n+1)!} = \\ = \sum_{m=0}^{n-1} {n+1\choose m} \frac{B_{m}}{(n+1)!} + {n+1 \choose n}\frac{B_n}{(n+1)!} = 0
\end{align}
for any $n > 0$. Thus
$$
(n+1)B_n = -\sum_{m=0}^{n-1} {n+1\choose m} \frac{B_{m}}{(n+1)!} 
$$
which implies result $(2)$ from your tutorial (reassign $n\to m$, $m\to k$).
Now for $n = 0$ we have from $(1)$
$$
\frac{B_0}{0!}\frac{1}{1!} = 1
$$
and thus $B_0 = 1$.
