It is well know that if we have two polynomials $f$ and $g$ such that $$f(x)=\sum_{i=0}^nf_ix^i,\quad g(x)=\sum_{j=0}^mg_jx^j$$ then the coefficient of $x^s$ in the product $f(x)\cdot g(x)$ is exactly $$\sum_{i+j=s}f_ig_j$$
But what about the product of $3$ polynomials? can we say, given another polynomial $$h(x)=\sum_{k=0}^rh_kx^k$$ the coefficient of $x^s$ in the product $fgh$ is $$\sum_{i+j+k=s}f_ig_jh_k$$ I tried to prove it like this, the coefficient of $x^s$ in $(fg)\cdot h$ is $$\sum_{i+j+k=s}\left(\sum_{i+j=u}f_ig_j\right)h_k$$ But I don't know how to proceed.
It's simpler if you write a polynomial in the form $f(x) = \sum_{i \ge 0} f_i x^i$ (with the underlying assumption that all $f_i$ are zero for $i > n$, but we only assume this $n$ exists, we don't need to explicit it). $$ (f \cdot g \cdot h)(x) = \left(\sum_{i \ge 0} f_i x^i \right) \left(\sum_{j \ge 0} g_j x^j \right) \left(\sum_{k \ge 0} h_k x^k \right) = \sum_{i,j,k \ge 0} f_ig_jh_k x^{i+j+k} = \sum_{\ell \ge 0} \left( \underset{i+j+k=\ell}{\sum_{i,j,k \ge 0}} (f_ig_jh_k) \right) x^{\ell}. $$ You can see how this easily generalizes to a product of $n$ factors.
Hope that helps,