On an expansion of $(1+a+a^2+\cdots+a^n)^2$ 
Question: What is an easy or efficient way to see or prove that
  $$
1+2a+3a^2+\cdots+na^{n-1}+(n+1)a^n+na^{n+1}+\cdots+3a^{2n-2}+2a^{2n-1}+a^{2n}\tag{1}
$$
  is equal to
  $$
(1+a+a^2+\cdots+a^n)^2\tag{2}
$$
  Maybe this is a particular case of a more general, well-known result?

Context: This is used with $a:=e^{it}$ to get an expression in terms of $\sin$ for the Fejér kernel.
Thoughts: I thought about calculating the coefficient $c_k$ of $a^k$. But my method is not so obvious that we can get from $(1)$ to $(2)$ in the blink of an eye.
$\mathbf{k=0}$ : clearly $c_0=1$.
$\mathbf{1\leq k\leq n}$ : $c_k$ is the number of integer solutions of $x_1+x_2=k$ with $0\leq x_1,x_2\leq k$, which in turn is the number of ways we can choose a bar $|$ in
$$
\underbrace{|\star|\star|\cdots|\star|}_{k\text{ stars}}
$$
So $c_k=k+1$.
$\mathbf{k=n+i\quad(1\leq i\leq n)}$ : $c_k$ is the number of integer solutions to $x_1+x_k=n+i$ with $0\leq x_1,x_2\leq n$, which in turn is the number of ways we can choose a bar $|$ in
$$
\underbrace{|\star|\star|\cdots|\star|}_{n+i\text{ stars}}
$$
different from the $i$-th one from each side. So $c_k=(n+i)+1-2i=n-i+1$.
 A: (Don't forget, the square eventually stops, and so you get a diamond shape at the end)

A: Just multiply as,
$$1(1+a+a^2...+a^n)$$
$$+$$
$$a(1+a+a^2...+a^n)$$
$$+$$
$$a^2(1+a+a^2...+a^n)$$
$$+.....$$
$$a^n(1+a+a^2...+a^n)$$
This gives the sum of,
$$\begin{align}1+\ \ a+&\ \ a^2+\ \ a^3+\dots+\ \ a^n
\\\ \ a+&\ \ a^2+\ \ a^3+\dots+\ \ a^n+\ \ a^{n+1}
\\&\ \ a^2+\ \ a^3+\dots+\ \ a^n+\ \ a^{n+1}+\ \ a^{n+2}
\\&\quad\ \ \ \ \ \ \ \ \ a^3+\dots+\ \ a^n+\ \ a^{n+1}+\ \ a^{n+2}+\ \ a^{n+3}\\&\ \ \ \qquad\qquad\vdots\qquad\qquad\qquad\qquad\qquad\qquad\qquad\vdots
\\\hline1+2a+&3a^2+4a^3+\dots+(n+1)a^n+na^{n+1}+\dots+a^{2n}\end{align}$$
A: Try to utilize a very well known formula: $$\left(\sum_i x_i\right)^2=\sum_i x_i^2+2\sum_{i<j}x_ix_j\,.$$ As you can see, I started from the end. But it does not matter for the proof.
A: Hint:
Use synthetic division twice after you you've rewritten the expression as 
$$\frac{(a^{n+1}-1)^2}{(a-1)^2}=\frac{a^{2n+2}-2a^{n+1}+1}{(a-1)^2}$$
$$\begin{array}{*{11}{r}}
&1&0&0&\dotsm&0&-2&0&0&\dots&0&0&1\\
&\downarrow&1&1&\dotsm&1&1&-1&-1&\dotsm&-1&-1&-1\\
\hline
\times1\quad&1&1&1&\dotsm&1&-1&-1&-1&\dotsm&-1&-1&0\\
&\downarrow&1&2&\dotsm&n&n+1&n&n-1&\dotsm&2&1\\
\hline
\times1\quad&1&2&3&\dotsm&n+1&n&n-1&n-2&\dotsm&1&0
\end{array}$$
