Let $n\geq 1,$ and let $f$ and $g$ have $n$th derivatives on $(a,b).$ Show that $(fg)^{(n)}(x)=\displaystyle\sum_{k=0}^n {n\choose k} f^{(n-k)}(x)g^{k}(x).$
My work. I think I can use induction to solve this.
Consider $n=1.$ Then $(fg)^{(1)}(x)=f'(x)g(x)+f(x)g'(x)$ and $\displaystyle\sum_{k=0}^1 {1\choose k} f^{(1-k)}(x)g^{(k)}(x)=f'(x)g(x)+f(x)g'(x).$ Hence the result is true for $n=1.$
Assume the result is true for some $m\in \mathbb{N}, 1\leq m< n.$
By assumption, $(fg)^{(m)}(x)=\displaystyle\sum_{k=0}^m{m\choose k}f^{(m-k)}(x)g^{(k)}(x).$ Since $(fg)^{(m+1)}(x)=\dfrac{d}{dx}\left[(fg)^{(m)}(x)\right],$ we have that $$(fg)^{(m+1)}(x)=\displaystyle\sum_{k=0}^m {m\choose k}f^{(m+1-k)}(x)g^{(k)}(x)+\displaystyle\sum_{k=0}^m{m\choose k}f^{(m-k)}(x)g^{(k+1)}(x)\\ =f^{(m+1)}(x)g(x)+\displaystyle\sum_{k=1}^m{m\choose k}f^{(m+1-k)}(x)g^{(k)}(x)+\displaystyle\sum_{k=0}^m {m\choose k}f^{(m-k)}(x)g^{(k+1)}(x)\\ =f^{(m+1)}(x)g(x)+\displaystyle\sum_{k=0}^{m-1}{m\choose k+1}f^{(m-k)}(x)g^{(k+1)}(x)+\displaystyle\sum_{k=0}^m {m\choose k}f^{(m-k)}(x)g^{(k+1)}(x)$$ (by summation index properties)
Now note that $${m\choose k+1}+{m\choose k} =\dfrac{m!}{(k+1)!(m-k-1)!}+\dfrac{m!}{k!(m-k)!}\\ =\dfrac{m!(m-k)}{(k+1)!(m-k-1)!}+\dfrac{m!(k+1)}{(k+1)!(m-k)!}\\ =\dfrac{(m+1)!}{(k+1)!(m-k)!}\\ ={m+1\choose k+1}.$$
So the sum is equivalent to $$f^{(m+1)}(x)g(x)+\displaystyle\sum_{k=0}^{m-1}{m+1 \choose k+1} f^{(m-k)}(x)g^{(k+1)}(x)+f(x)g^{(m+1)}(x)\\ =f^{(m+1)}(x)g(x)+\displaystyle\sum_{k=1}^{m} {m+1 \choose k} f^{(m+1-k)}(x)g^{(k)}(x)+f(x)g^{(m+1)}(x)$$ (by summation properties) $$=\displaystyle\sum_{k=0}^{m+1} {m+1\choose k} f^{(m+1-k)}(x)g^{(k)}(x),$$ and so the result holds for all $n\in\mathbb{N}.$
This argument is a little long, so I guess it can be shortened.