prove three claims involving the derivative of a function 
Prove that $(x\dfrac{d}{dx} )^m (1-x)^n $ evaluated at $x=1$ is $\sum_{i=1}^n {n\choose i}(-1)^i i^m .$


Prove that all terms of $(x\dfrac{d}{dx} )^m (1-x)^n $ have a nonzero factor of $1-x$ when $m<n$.


Prove that when $m=n$, the only nonzero term is $(-1)^n x^n n!$.

I think induction can be used to prove all of the above claims, but I need to state the inductive hypothesis carefully (in particular it needs to be strong enough to prove the inductive step). It seems likely that the exact formula for $(x\dfrac{d}{dx} )^m (1-x)^n $ will probably be too complicated to express; it might be easier to say it equals $(1-x)p_m(x) + q_m(x)$ for some polynomials $p_m(x)$ and $q_m(x)$. $p(x)$ likely has degree $n-m$. It should then be possible (though not necessarily easy) to show $\sum_{i=1}^n {n\choose i}(-1)^i i^m = 0$ for all $m < n.$
 A: The operator $xD = x\frac{d}{dx}$ acts on the monomial $x^k$ (with $k\geq 1$) by bringing it into $k x^k$, so $(xD)^m x^k = k^m x^k$. Since
$$ (1-x)^n = \sum_{k=0}^{N}\binom{n}{k}(-1)^k x^k $$
by the binomial theorem, we have
$$ (xD)^m (1-x)^n = \sum_{k=1}^{n}\binom{n}{k}(-1)^k k^m x^k \tag{1}$$
and the evaluation at $x=1$ leaves us with
$$ \left.(xD)^m(1-x)^n\right|_{x=1} = \sum_{k=1}^{n}\binom{n}{k}(-1)^k k^m. \tag{2}$$
If $m<n$, in order to prove that the RHS of $(1)$ is a multiple of $(1-x)$ it is enough to check that its evaluation at $x=1$ equals zero. Notice that the RHS of $(2)$, in terms of the forward difference operator $\Delta$, is $\Delta^n x^m$ evaluated at $x=0$. Since $\deg(\Delta p) = \deg p-1$ for any non-constant polynomial, it is pretty straightforward that the RHS of $(2)$ equals zero if $m<n$. On the other hand $\Delta^n x^n=n!$ since the leading coefficient of $\Delta p$ is the same as the leading coefficient of $p'$.
Why? Here's why. $(\Delta p)(x)=p(x+1)-p(x)$, $(\Delta^2 p)(x)=p(x+2)-2p(x+1)+p(x)$, $(\Delta^3 p)(x)=p(x+3)-3p(x+2)+3p(x+1)-p(x)$ and in general
$$(\Delta^n p)(x) = \sum_{k=0}^{n}\binom{n}{k}(-1)^n p(x+k)$$
so by considering $p(x)=x^m$ we have
$$(\Delta^n p)(x) = \sum_{k=0}^{n}\binom{n}{k}(-1)^n (x+k)^m $$
and the evaluation at $x=0$ gives the RHS of $(2)$. $\Delta$ is a linear operator and since $\Delta$ applied to $x^d$ (with $d\geq 1$) gives $d x^{d-1}+\ldots$, $\Delta$ applied to any non-constant polynomial $p$ gives a polynomial whose leading coefficient is the same as $p'$. It follows that $\Delta^n x^n$ is a constant which equals $\frac{d^n}{dx^n} x^n = n!$ and $\Delta^n x^m$ is just the constant zero if $n>m$.
Alternative approach. The operator $xD$ is conjugated to $D$ in the following sense. In order to apply $xD$ to $f(x)$, it is enough to

*

*Replace $x$ with $e^z$

*Differentiate with respect to $z$

*Replace back $z$ with $\ln x$
In formula $(xD)p(x) = \left. \frac{d}{dz} p(e^z)\right|_{z=\ln x}$, so
$$(xD)^m p(x) = \left. \frac{d^m}{dz^m} p(e^z)\right|_{z=\ln x}\tag{3}$$
If we start with $p(x)=(x-1)^n$ we have that $(e^z-1)^n$ is an entire function with zeroes of multiplicity $n$ at the elements of $2\pi i\mathbb{Z}$. The application of $\frac{d}{dz}$ decreases the multiplicity of any zero by one, so if $m<n$ we have that $\frac{d^m}{dz^m}(e^z-1)^n$ has zeroes of multiplicity $n-m\geq 1$ at the elements of $2\pi i\mathbb{Z}$. This gives that $(e^z-1)$ is a factor of $\frac{d^m}{dz^m}(e^z-1)^n$ and $(x-1)$ is a factor of $(xD)^m(x-1)^n$.
