Solutions

Let's prove Leibniz's formula by induction. For the base case $n=0$, we have

$$(f\cdot g)^{(0)}(a)=f(a)g(a)=\binom{0}{0}f^{(0)}(a)g^{(0)}(a).$$

For the induction step, suppose the formula holds for $n$. Then,

$$\begin{align*} (f\cdot g)^{(n+1)}(a)&=((f\cdot g)^{(n)}(a))' \\ &=\left(\sum_{k=0}^n\binom{n}{k}f^{(k)}(a)g^{(n-k)}(a)\right)' \\ &=\sum_{k=0}^n\binom{n}{k}\left(f^{(k+1)}(a)g^{(n-k)}(a)+f^{(k)}(a)g^{(n-k+1)}(a)\right) \\ &=\sum_{k=0}^n\binom{n}{k}f^{(k+1)}g^{(n-k)}+\sum_{k=0}^n\binom{n}{k}f^{(k)}g^{(n+1-k)} \\ &=f^{(0)}g^{(n+1)}+f^{(n+1)}g^{(0)}+\sum_{k=1}^n\binom{n}{k-1}f^{(k)}g^{(n+1-k)}+\sum_{k=1}^n\binom{n}{k}f^{(k)}g^{(n+1-k)} \\ &=f^{(0)}g^{(n+1)}+f^{(n+1)}g^{(0)}+\sum_{k=1}^n f^{(k)}g^{(n+1-k)}\left(\binom{n}{k-1}+\binom{n}{k}\right) \\ &=f^{(0)}g^{(n+1)}+f^{(n+1)}g^{(0)}+\sum_{k=1}^n\binom{n+1}{k}f^{(k)}g^{(n+1-k)} \\ &=\sum_{k=0}^{n+1}\binom{n+1}{k}f^{(k)}g^{(n+1-k)}. \end{align*}$$

with

$$\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$$

because

$$\begin{align*} \binom{n}{k-1}+\binom{n}{k}&=\frac{n!}{(k-1)!(n-k+1)!}+\frac{n!}{k!(n-k)!} \\ &=\frac{n!k+n!(n+1-k)}{k!(n+1-k)!} \\ &=\frac{n!(k+n+1-k)}{k!(n+1-k)!} \\ &=\frac{(n+1)!}{k!(n+1-k)!}. \end{align*}$$