I came across a notation that was unfamiliar (Bernoulli Numbers) that I reproduce below
$$ \sum_{j=0}^{n} \left(\begin{array}{cl} n+1\\ j\\ \end{array}\right) B_j = 0 \hspace{1cm} with \hspace{1cm} B_0 = 1$$
I'm not sure how to approach this notation, nor how to use it to produce the Bernoulli numbers. Also, as more of a bonus, the authors of the source say that the formula can be reproduced from
$$ \frac{x}{e^x-1} = \sum_{n=0}^{\infty}B_n\frac{x^n}{n!} \hspace{1cm} |x|<2\pi $$
"by multiplying both sides by $e^x−1$, using the Cauchy product with the Maclaurin series for $e^x−1$, and then equating the coefficients of the powers of $x$."
The given identity allows you to compute the Bernoulli number $B_n$ for $n\geq 1$ recursively. Note that $\binom{n+1}{j}$ is not a vector, it is an integer number called binomial coefficient.
By isolating the last term of the sum we get the formula: $$B_n=-\frac{1}{n+1}\sum_{j=0}^{n-1}\binom{n+1}{j}B_j.$$ For example, since $B_0=1$, it follows that $$\begin{align} B_1&=-\frac{1}{1+1}\sum_{j=0}^{1-1}\binom{1+1}{j}B_j=-\frac{B_0}{2}=-\frac{1}{2},\\ B_2&=-\frac{1}{2+1}\sum_{j=0}^{2-1}\binom{2+1}{j}B_j=-\frac{B_0}{3}-\frac{3B_1}{3}=\frac{1}{6}. \end{align}$$ Are you able to compute $B_3$?
As regards a proof of exponential generating functions of the Bernoulli numbers, you may take a look HERE.