I need to algebraically, or using basic calculus, show that $$\displaystyle\frac{d^m}{dx^m}[f(x)g(x)] + {{m+1}\choose k}\frac{d^{m+1+k}}{dx^{m+1+k}}f(x)\frac{d^k}{dx^k}g(x) = \frac{d^{m+1}}{dx^{m+1}}[f(x)g(x)]$$
For reference, it's part of a proof by induction to prove that $\displaystyle\frac{d^n}{dx^n}[f(x)g(x)] = \sum\limits_{k=0}^n {n\choose k}\frac{d^{n-k}}{dx^{n-k}}f(x)\frac{d^k}{dx^k}g(x)$
Any objections to using induction is also welcome.
You can prove the inductive step as follows: knowing that $$\frac{d^n}{dx^n}[f(x)g(x)] = \sum\limits_{k=0}^n {n\choose k}\frac{d^{n-k}}{dx^{n-k}}f(x)\frac{d^k}{dx^k}g(x)$$ you have $$\frac{d^{n+1}}{dx^{n+1}}[f(x)g(x)] =\frac{d}{dx}\sum\limits_{k=0}^n {n\choose k}\frac{d^{n-k}}{dx^{n-k}}f(x)\frac{d^k}{dx^k}g(x)$$ $$ \sum\limits_{k=0}^n\left[ {n\choose k}\frac{d^{n-k+1}}{dx^{n-k+1}}f(x)\frac{d^k}{dx^k}g(x)+ {n\choose k}\frac{d^{n-k}}{dx^{n-k}}f(x)\frac{d^{k+1}}{dx^{k+1}}g(x)\right]=$$ $$ =\sum\limits_{k=0}^n {n\choose k}\frac{d^{n-k+1}}{dx^{n-k+1}}f(x)\frac{d^k}{dx^k}g(x)+ \sum\limits_{k=0}^n {n\choose k}\frac{d^{n-k}}{dx^{n-k}}f(x)\frac{d^{k+1}}{dx^{k+1}}g(x)=$$ $$ =\sum\limits_{k=0}^n {n\choose k}\frac{d^{n-k+1}}{dx^{n-k+1}}f(x)\frac{d^k}{dx^k}g(x)+ \sum\limits_{k=1}^{n+1} {n\choose {k-1}}\frac{d^{n-k+1}}{dx^{n-k+1}}f(x)\frac{d^{k}}{dx^{k}}g(x)= $$ $$ =\frac{d^{n+1}}{dx^{n+1}}f(x)\ g(x)+\sum\limits_{k=1}^n\left[ {n\choose k}+{n\choose {k-1}}\right]\frac{d^{n-k+1}}{dx^{n-k+1}}f(x)\frac{d^k}{dx^k}g(x)+ f(x)\frac{d^{n+1}}{dx^{n+1}}g(x)\ . $$ Now you just have to rewrite: $${n\choose k}+{n\choose {k-1}}=\frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}=\frac{n!(n-k+1+k)}{k!(n-k+1)!}=$$$$=\frac{(n+1)!}{k!(n-k+1)!}={{n+1}\choose {k}}\ ,$$ so finally your sum becomes: $$ \frac{d^{n+1}}{dx^{n+1}}f(x)\ g(x)+\sum\limits_{k=1}^n{{n+1}\choose {k}}\frac{d^{n-k+1}}{dx^{n-k+1}}f(x)\frac{d^k}{dx^k}g(x)+ f(x)\frac{d^{n+1}}{dx^{n+1}}g(x)=$$$$=\sum\limits_{k=0}^{n+1}{{n+1}\choose {k}}\frac{d^{n-k+1}}{dx^{n-k+1}}f(x)\frac{d^k}{dx^k}g(x)\ . $$
