Show that
$\left(\sum\limits_{j=0}^n a_jx^j\right)\left(\sum\limits_{k=0}^n b_kx^k\right)=\sum\limits_{m=0}^{2n}\left(\sum\limits_{j+k=m} a_jb_k\right)x^m=\sum\limits_{m=0}^{2n}(a_0b_m+a_1b_{m-1}+...+a_mb_0)x^m $
If $a_k=b_k=:0$ for $k=n+1,...,2n$
$\sum\limits_{m=0}^{2n}\left(\sum\limits_{j+k=m} a_jb_k\right)x^m=a_0b_0x^0+(a_1b_0+a_0b_1)x^1+...+(a_{0}b_{2n-1}+a_{1}b_{2n-2}+...+a_{2n-2}b_{1}+a_{2n-1}b_0)x^{2n-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+(a_{0}b_{2n}+a_{1}b_{2n-1}+...+a_{2n-1}b_{1}+a_{2n}b_{0})x^{2n}=\sum\limits_{m=0}^{2n}(a_0b_m+a_1b_{m-1}+...+a_mb_0)x^m$
Since $a_k=b_k=:0$ for $k=n+1,...,2n$
$=a_0b_0x^0+(a_1b_0+a_0b_1)x^1+...+(a_{0}b_{n}+a_{1}b_{n-1}+...+a_{n-1}b_{1}+a_{n}b_{0})x^{n}+(a_{1}b_{n}+a_{2}b_{n-1}+...+a_{n-1}b_{2}+a_nb_{1})x^{n+1}+$
$...+(a_{n-2}b_{n}+a_{n-1}b_{n-1}+a_{n}b_{n-2})x^{2n-2}+(a_{n-1}b_{n}+a_{n}b_{n-1})x^{2n-1}+a_nb_nx^{2n}$
$=(a_nx^n+a_{n-1}x^{n-1}+...+a_2x^2+a_1x^1+a_0x^0)(b_nx^n+b_{n-1}x^{n-1}+...+b_2x^2+b_1x^1+b_0x^0)$
$=\left(\sum\limits_{j=0}^n a_jx^j\right)\left(\sum\limits_{k=0}^n b_kx^k\right)$
Is this what the book wanted from me? Thanks for feedback :)
While technically correct, really you've skipped over the entire proof. Although it is "obvious" that you can redistribute variables in the way you've done on the second-last equality, this is exactly what the question is asking you to prove, so you should definitely be more rigorous about it.
As a sort of intuitive "litmus test" for whether or not proofs of results like this are rigorous enough: if you have to use a "..." in your proof to make it convincing, it's probably not.
To prove this result, I would probably use mathematical induction on $n$. Start with the $n = 0$ base case and then prove the inductive step using basic facts about sums (more basic than the fact that you are being asked to prove), such as the fact that $$ \sum_{k=0}^{n} a_k = \left(\sum_{k=0}^{n-1} a_k\right) + a_n. $$ This should lead you to a "...-free" proof.