Here’s one way of looking at it.
Suppose that you have $f(x)=\sum_{n\ge 0}a_nx^n\;,$ and you multiply both sides by $\frac1{1-x}$:
$\begin{align*} \left(\frac1{1-x}\right)f(x)&=\left(\sum_{n\ge 0}x^n\right)\left(\sum_{n\ge 0}a_nx^n\right)\\ &=(1+x+x^2+\dots)(a_0+a_1x+a_2x^2+\dots)\\ &=a_0 +(a_0+a_1)x+(a_0+a_1+a_2)x^2+\dots\\ &=\sum_{n\ge }\left(\sum_{k=0}^na_k\right)x^n\;. \end{align*}$
In other words, the coefficient of $x^n$ in the product is just $a_0+a_1+\dots+a_n$.
Now think about the construction of Pascal’s triangle:
$\begin{array}{c} 1\\ 1&1\\ 1&2&1\\ 1&3&3&1\\ 1&4&6&4&1\\ 1&5&10&10&5&1\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots \end{array}$
The binomial coefficient $\binom{n}k$ is the entry in row $n$, column $k$ (numbered from $0$). Moreover, since $\binom{n}k=\binom{n-1}k+\binom{n-1}{k-1}\;,$ each entry is the sum of the numbers above it in the column immediately to the left: this is the identity $\sum_{i=0}^n\binom{i}k=\binom{n+1}{k+1}\;.\tag{1}$
In the first ($k=0$) column of Pascal’s triangle you have the coefficients in the power series expansion of $\frac1{1-x}$. We saw above that the coefficients in the power series expansion of $\frac1{(1-x)^2}$ are just the cumulative sums of these coefficients, $1,2,3,\dots$, but these are just the entries in the second ($k=1$) column of Pascal’s triangle. Similarly, the coefficients in the power series expansion of $\frac1{(1-x)^3}$ are the cumulative sums of $1,2,3\dots$, or $1,3,6,\dots$, the numbers in the third ($k=2$) column of Pascal’s triangle. In general, the coefficients in the power series expansion of $\frac1{(1-x)^{k+1}}$ must be the binomial coefficients in the $k$ column of Pascal’s triangle, those of the form $\binom{n}k$. All that remains is to get the row indexing right: we want the $1$ that is the first non-zero entry in column $k$ to be the constant term. It’s in row $k$, so the coefficient of $x^n$ must in general be the binomial coefficient in row $n+k$, and we get
$\frac1{(1-x)^{k+1}}=\sum_{n\ge 0}\binom{n+k}kx^n=\sum_{n\ge 0}\binom{n+k}nx^n\;.$