If $$f(x)=a_nx^n+...+a_0$$ and $$g(x)=b_mx^m+…+b_0$$ then $$f(x)\cdot g(x)=c_{m+n}x^{m+n}+...+c_0$$ where $c_k=\sum_{r+s=k}a_rb_s,\quad k=0,....,m+n$
I know that the degree $0$ and $(m+n)$ exists, and I also understand the Formula for those cases.
Now if I pick a degree $k$ in between $0$ and $(m+n)$. Then if I always take a summand of the first factor $f(x)$ i. e. $\, a_i$, then $i$ cannot be bigger than $k$ because I would have to multiply it with a monomonial with at least degree $0$.
Those are my thoughts so far; how can I continue this chain of thoughts in order to prove the Formula for the coefficients, i. e. why does it have to be this Formula?
I've been playing around with trying to find a more visual way of proving the Cauchy Product, which amongst other things, multiplies two polynomials together as you've asked for;
My starting point was thinking about a specific example; $$(2+7x+3x^2)(6+4x+5x^2) =$$
\begin{array}{c|c|c|c} &2 &7x &3x^2 \\ \hline 6 &12&42x&18x^2\\ \hline 4x &8x&28x^2&12x^3\\ \hline 5x^2 &10x^2&35x^3&15x^4\\ \hline \end{array} So
$$(2+7x+3x^2)(6+4x+5x^2) = 12+50x+56x^2+47x^3+15x^4$$
Generalising this, $$(a_0+a_1x+a_2x^2+a_3x^3+\dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+\dots+b_mx^m)=$$ \begin{array}{c|c|c|c|c|c|c} &a_0 &a_1x &a_2x^2 &a_3x^3&\dots&a_nx^n \\ \hline b_0 &a_0b_0&a_1b_0x&a_2b_0x^2&a_3b_0x^3&\dots&a_nb_0x^n\\ \hline b_1x &a_0b_1x&a_1b_1x^2&a_2b_1x^3&a_3b_1x^4&\dots&a_nb_1x^{n+1}\\ \hline b_2x^2 &a_0b_2x^2&a_1b_2x^3&a_2b_2x^4&a_3b_2x^5&\dots&a_nb_2x^{n+2}\\ \hline b_3x^3&a_0b_3x^3&a_1b_3x^4&a_2b_3x^5&a_3b_3x^6&\dots&a_nb_3x^{n+3}\\ \hline \dots&\dots&\dots&\dots&\dots&\dots&\dots\\ \hline b_mx^m&a_0b_mx^m&a_1b_mx^{m+1}&a_2b_mx^{m+2}&a_3b_mx^{m+3}&\dots&a_nb_mx^{n+m}\\ \hline \end{array} So, $$(a_0+a_1x+a_2x^2+a_3x^3+\dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+\dots+b_mx^m)$$ $$=a_0b_0$$ $$+(a_0b_1+a_1b_0)x$$ $$+(a_0b_2+a_1b_1+a_2b_0)x^2$$ $$+(a_0b_3+a_1b_2+a_2b_1+a_3b_0)x^3$$ $$+\dots$$ $$+a_nb_nx^{n+m}$$ which is what you were trying to prove.