Generating Function for Bernoulli Polynomials proof Generating Function for $B_n$ is:
$\displaystyle \frac {t e^{t x} } {e^t - 1} = \sum_{k \mathop = 0}^\infty \frac { {B_k} x} {k!} t^k$

By definition of the generating function for Bernoulli numbers:
$\frac t {e^t - 1} = \sum_{k \mathop = 0}^\infty \frac {B_k} {k!} t^k $

By Power Series Expansion for Exponential Function:
$e^{tx}= \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} t^k $

Thus:
$\color{green}{\frac {t e^{t x} } {e^t - 1}= ( {\sum_{k \mathop = 0}^\infty \frac {B_k} {k!} t^k} )({\sum_{k \mathop = 0}^\infty \frac {x^k} {k!} t^k})}$

We know that:

$F(x)=f_0+f_1x+f_2x^2+...$
$G(x)=g_0+g_1x+g_2x^2+...$
$F(x)=G(x)\  \Longleftrightarrow \ f_0=g_0, \ f_1=g_1, \ f_2=g_2, \ ...$
$\alpha * F(x)+\beta *G(x)=\displaystyle\sum_{n =0 }^{\infty}(\alpha * f_n +\beta * g_n)x^n=$
$.\quad \quad \quad \quad \quad \quad  \quad \quad \quad \quad \quad(\alpha * f_0 +\beta * g_0)+(\alpha * f_1 +\beta * g_1)+(\alpha * f_2+ \beta * g_2)+...$
$\color{blue}{F(x)*G(x)=\displaystyle\sum_{n =0 }^{\infty}(\displaystyle\sum_{k =0 }^{n}f_kg_{n-k})x^n=}$
$\color{blue}{.\quad \quad \quad f_0g_0+(f_0g_1+f_1g_0)x}$
$\color{blue}{.\quad \quad \quad +(f_0g_2+f_1g_1+f_2g_0)x^2}$
$\color{blue}{.\quad \quad \quad +(f_0g_3+f_1g_2+f_2g_1+f_3g_0)x^3+...}$

I don't not understand how combining like powers worked out below? How was it calculated that in red below?:

$\color{green}{\frac {t e^{t x} } {e^t - 1} }= \sum_{k \mathop = 0}^\infty \color{red}{t^k}\color{black} \sum_{m \mathop = 0}^k \frac {B_{k - m} } {(k - m)!} \frac {x^m} {m!}$

$=\sum_{k \mathop = 0}^\infty \frac {t^k} {k!} \sum_{m \mathop = 0}^k \frac {k!} {( {k - m})! m!} B_{k - m} x^m$

$=\sum_{k \mathop = 0}^\infty \frac {t^k} {k!} \sum_{m \mathop = 0}^k \binom k m B_{k - m} x^m $
$=\sum_{k \mathop = 0}^\infty \frac {{B_k} (x)} {k!} t^k$

$\color{blue}{How \ that\ combining\ of\ same\ powers\ \ in \ red\ was\ done?\ Under\ what\ rules?}$

$\color{blue}{Am \ I \ right \ thinking \ that \ is \ case \ below?}$

$\color{green}{\frac {t e^{t x} } {e^t - 1}= \sum_{k \mathop = 0}^\infty \sum_{m \mathop = 0}^k \frac {B_{k - m} } {(k - m)!} \frac {x^m} {m!}\color{red}{t^m*t^{k-m}}}$
 A: What you are doing is called the binomial convolution of two exponential generating functions.
Let $$f(x) = \sum_{k=0}^\infty \frac{a_k}{k!} x^k, \\ g(x) = \sum_{m=0}^\infty \frac{b_m}{m!}x^m$$ be the exponential generating functions of the sequences $\{a_k\}_{k \ge 0}$ and $\{b_m\}_{m \ge 0}$.  Then $$f(x) g(x) = \sum_{k=0}^\infty \sum_{m=0}^\infty \frac{a_k}{k!} \frac{b_m}{m!} x^k x^m,$$ where the double sum is taken over all nonnegative pairs of integers $(k,m)$.  We want to express this as a sum over a single index variable; i.e., to determine the coefficients $\{c_n\}_{n \ge 0}$ satisfying $$f(x) g(x) = \sum_{n=0}^\infty \frac{c_n}{n!} x^n.$$  To this end, we note that for some fixed $n \ge 0$, the coefficient of the $x^n$ term is the sum over all $(k,m)$ such that $k + m = n$.  In other words, we can write this as $$f(x) g(x) = \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{a_k}{k!} \frac{b_{m-k}}{(m-k)!}\right) x^k x^{n-k} = \sum_{n=0}^\infty \left(\sum_{k=0}^n \binom{n}{k} a_k b_{n-k} \right) \frac{1}{n!} x^n.$$  Therefore, $$c_n = \sum_{k=0}^n \binom{n}{k} a_k b_{n-k}$$ is the desired relationship between the EGFs of $f$ and $g$.
To understand this on a concrete level, consider the set of nonnegative integer lattice points $(k,m)$, which we can write in a table like this:
$$\begin{array}{c|ccccc}
(k,m) & 0 & 1 & 2 & 3 & 4 & \cdots \\
\hline
0 & (0,0) & (0,1) & (0,2) & (0,3) & \color{red}{(0,4)} & \cdots \\
1 & (1,0) & (1,1) & (1,2) & \color{red}{(1,3)} & (1,4) & \cdots \\
2 & (2,0) & (2,1) & \color{red}{(2,2)} & (2,3) & (2,4) & \cdots \\
3 & (3,0) & \color{red}{(3,1)} & (3,2) & (3,3) & (3,4) & \cdots \\
4 & \color{red}{(4,0)} & (4,1) & (4,2) & (4,3) & (4,4) & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots 
\end{array}
$$
When we pick an $n$, we are picking a diagonal of the array such that the sum of the coordinates is $n$.  Above, the choice $n = 4$ is highlighted in red.  Then for each choice of $k$ ranging from $0$ to $n$ within this diagonal, there is a unique choice $m = n-k$.  So whereas the original sum added the array row by row, the rearranged sum adds the array diagonal by diagonal.
