proof of a function times the nth derivative of a function formula 
Prove by induction or otherwise that when u and v are functions of x $$vD^nu=D^n(uv)-n{D}^{n-1}(uDv)+\frac{n(n-1)D^{n-2}(uD^2v)}{2}-...$$an question from differential calculus by Ferrar


so far I have checked so base case n=1 and $$vDu=D(v)u-D(v)u+vD(u)=D(v)u+vD(u)-D(v)u=D(uv)-D^{0}(uDV)$$
so assuming for some k $$vD^ku=D^k(uv)-k{D}^{k-1}(uDv)+\frac{k(k-1)D^{k-2}(uD^2v)}{2}-...$$
note by libniz formula for the nth derivate formula $$D^k(uv)=\sum_{r=0}^k \binom{k}{r}D^{r}vD^{n-r}u=vD^{k}u+\sum_{r=1}^{k} \binom{k}{r}D^{r}vD^{n-r}u$$ Hence all other terms aside from the first go to 0 when combined
$\implies$ $$\sum_{r=1}^{k} \binom{k}{r}D^{r}vD^{n-r}u+-k{D}^{k-1}(uDv)+\frac{k(k-1)D^{k-2}(uD^2v)}{2}-...=0=(i)$$
now if we consider the k+1th case to be true then we need show that $$vD^{k+1}u=D^{k+1}(uv)-{k+1}{D}^{k}(uDv)+\frac{(k+1)(k)D^{k-1}(uD^2v)}{2}-...$$ $=$in an annalgous manner$$\sum_{r=1}^{k+1} \binom{k+1}{r}D^{r}vD^{n-r}u+-(k+1){D}^{k}(uDv)+\frac{(k+1)(k)D^{k-1}(uD^2v)}{2}-...=0=(ii)$$ so if i can show from (i) that (ii) is true then I will be done if any one has any help to offer with this solution please answer or if anyone see's a more elegant solution feel free to answer, or see's any mistakes in my working
 A: I will use $f$ and $g$ instead of $u$ and $v$.
By Leibniz formula :
$$\forall k \in \{0, \ldots, n - 1\}, \left(f g^{(k)}\right)^{(n - k)} = \sum_{p = 0}^{n - k} \binom{n - k}{p} f^{(p)} \left(g^{(k)}\right)^{(n - k - p)} = \sum_{p = 0}^{n - k} \binom{n - k}{p} f^{(p)} g^{(n - p)}$$
then :
$$\begin{array}{lcl}
\displaystyle \sum_{k = 0}^{n - 1} (-1)^k \binom{n}{k} \left(f g^{(k)}\right)^{(n - k)} & = & \displaystyle \sum_{k = 0}^n (-1)^k \binom{n}{k} \sum_{p = 0}^{n - k} \binom{n - k}{p} f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{k = 0}^{n - 1} \sum_{p = 0}^{n - k} (-1)^k \binom{n}{k} \binom{n - k}{p} f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{p = 1}^n \left(\sum_{k = 0}^{n - p} (-1)^k \binom{n}{k} \binom{n - k}{p}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{p = 1}^n \left(\sum_{k = 0}^{n - p} (-1)^k \dfrac{n!}{k! (n - k)!} \dfrac{(n - k)!}{p! (n - k - p)!}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{p = 1}^n \left(\sum_{k = 0}^{n - p} (-1)^k \dfrac{n!}{(k + p)! (n - k - p)!} \dfrac{(k + p)!}{p! k!}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{p = 1}^n \left(\sum_{k = 0}^{n - p} (-1)^k \binom{n}{k + p} \binom{k + p}{p}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{p = 1}^n \left(\sum_{k = p}^n (-1)^{k - p} \binom{n}{k} \binom{k}{p}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{p = 1}^n \left(\sum_{k = p}^n (-1)^{k - p} \dfrac{n!}{k! (n - k)!} \dfrac{k!}{p! (k - p)!}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{p = 1}^n \left(\sum_{k = p}^n (-1)^{k - p} \dfrac{n!}{p! (n - p)!} \dfrac{(n - p)!}{(n - k)! (k - p)!}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{p = 1}^n \left(\sum_{k = p}^n (-1)^{k - p} \binom{n}{p} \binom{n - p}{k - p}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{p = 1}^n \binom{n}{p} \left(\sum_{k = p}^n (-1)^{k - p} \binom{n - p}{k - p}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle \sum_{p = 1}^n \binom{n}{p} \left(\sum_{k = 0}^{n - p} (-1)^k \binom{n - p}{k}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle f^{(n)} g^{(n - n)} \sum_{k = 0}^{n - n} (-1)^k \binom{n - n}{k} + \sum_{p = 1}^{n - 1} \binom{n}{p} \left(\sum_{k = 0}^{n - p} (-1)^k \binom{n - p}{k}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle f^{(n)} g + \sum_{p = 1}^{n - 1} \binom{n}{p} \left(\sum_{k = 0}^{n - p} (-1)^k \binom{n - p}{k}\right) f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle f^{(n)} g + \sum_{p = 1}^{n - 1} \binom{n}{p} \left(1 - 1\right)^{n - p} f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle f^{(n)} g + \sum_{p = 1}^n \binom{n}{p} 0 f^{(p)} g^{(n - p)} \\[3mm]
& = & \displaystyle f^{(n)} g
\end{array}$$
A: This is just a simplification of an already given answer (+1). We show the following is valid for $f,g$ at least $n$ times differentiable functions:
\begin{align*}
\color{blue}{\sum_{k=0}^n(-1)^k\binom{n}{k}\left(f\cdot g^{(k)}\right)^{(n-k)}=f^{(n)}g}\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k\binom{n}{k}\left(f\cdot g^{(k)}\right)^{(n-k)}}\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\sum_{p=0}^{n-k}\binom{n-k}{p}f^{(p)}g^{(n-p)}\tag{1.1}\\
&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\sum_{p=0}^{k}\binom{k}{p}f^{(p)}g^{(n-p)}\tag{1.2}\\
&=\sum_{p=0}^nf^{(p)}g^{(n-p)}\sum_{k=p}^n(-1)^{n-k}\binom{n}{k}\binom{k}{p}\tag{1.3}\\
&=\sum_{p=0}^n\binom{n}{p}f^{(p)}g^{(n-p)}\sum_{k=p}^{n}(-1)^{n-k}\binom{n-p}{k-p}\tag{1.4}\\
&=\sum_{p=0}^n\binom{n}{p}f^{(p)}g^{(n-p)}\sum_{k=0}^{n-p}(-1)^{n-p-k}\binom{n-p}{k}\tag{1.5}\\
&=\sum_{p=0}^n\binom{n}{p}f^{(p)}g^{(n-p)}(1-1)^{n-p}\tag{1.6}\\
&\,\,\color{blue}{=f^{(n)}g}
\end{align*}
and the claim (1) follows.

Comment:

*

*In (1.1) we apply the Leibniz product rule.


*In (1.2) we change the order of summation of the outer sum $k\to n-k$.


*In (1.3) we exchange the sums noting
\begin{align*}
\sum_{k=0}^n\sum_{p=0}^k\Box=\sum_{0\leq p\leq k\leq n}\Box=\sum_{p=0}^n\sum_{k=p}^n\Box
\end{align*}


*In (1.4) we use the binomial identity $\binom{n}{k}\binom{k}{p}=\binom{n}{p}\binom{n-p}{k-p}$.


*In (1.5) we shift the index of the inner sum to start with $k=0$.


*In (1.6) we apply the binomial theorem and observe $n=p$ only contributes to the sum.
