How to find sum of coefficients of $\frac{d^n}{dx^n} \left( Z(x)^m \right)$ I noticed something interesting  result but I do not know how to prove it (or disprove)
Function $U$ defined as  
$$ U(Z(x),Z'(x),Z''(x),Z'''(x),...,Z^{(n)}(x))=\frac{d^n}{dx^n} \left( Z^m(x) \right)$$ 
and $m,n>0$ 
I conjecture that
$$U(1,1,1,1,....,1)=m^n$$
Some examples

$n=2,m=3$
$$ U(Z(x),Z'(x),Z''(x))=\frac{d^2}{dx^2} \left( Z^3(x) \right)=6Z(x)Z'^2(x)+3Z^2(x)Z''(x)$$ 
$$ U(1,1,1)=9=3^2$$ 

$n=3,m=4$
$$ U(Z(x),Z'(x),Z''(x),Z'''(x))=\frac{d^3}{dx^3} \left( Z^4(x) \right)=24Z(x)Z'^3(x)+36Z^2(x)Z'(x)Z''(x)+4Z^3(x)Z'''(x)$$ 
$$ U(1,1,1,1)=64=4^3$$ 
I do not know how to prove or disprove my claim.
Could you please help me ?
Thanks
EDIT: I proved my claim for $n<4$ and for any $m$ 
$n=2$
$$ U(Z(x),Z'(x),Z''(x))=\frac{d^2}{dx^2} \left( Z^m(x) \right)=m(m-1)Z^{m-2}(x)Z'^2(x)+mZ^{m-1}(x)Z''(x)$$ 
$$ U(1,1,1)=m^2-m+m=m^2$$
$n=3$
$$ \frac{d^3}{dx^3} \left( Z^m(x) \right)=\frac{d}{dx} \left( m(m-1)Z^{m-2}(x)Z'^2(x)+mZ^{m-1}(x)Z''(x) \right)=m(m-1)(m-2)Z^{m-3}(x)Z'^3(x) +2m(m-1)Z^{m-2}(x)Z'(x)Z''(x)+m(m-1)Z^{m-2}(x)Z'(x)Z''(x)+mZ^{m-1}(x)Z'''(x)$$ 
$$ U(1,1,1,1)=m(m-1)(m-2)+3m(m-1)+m=m^3-3m^2+2m+3m^2-3m+m=m^3$$
But I do not know yet how to prove the general case.
 A: Here is a simple approach: for every fixed $(m,n)$, Leibniz rule shows that indeed there exists some unique (polynomial) function $U_{m,n}$ such that the desired identity holds for every regular enough function $Z$ and every $x$. Let us use a specific function $Z$, say, $$Z(x)=\mathrm e^x.$$ Then $Z^{(k)}(x)=Z(x)$ for every $k$ and $Z(x)^m=\mathrm e^{mx}$, hence $$U_{m,n}(\mathrm e^x,\mathrm e^x,\ldots,\mathrm e^x)=\frac{\mathrm d^n}{\mathrm dx^n}\left(\mathrm e^{mx}\right)=m^n\mathrm e^{mx},$$ in particular, for $x=0$, $\mathrm e^x=\mathrm e^{mx}=1$ hence 
$$U_{m,n}(1,1,\ldots,1)=m^n.$$ 
Likewise, for every $a$,
$$U_{m,n}(a,a,\ldots,a)=m^na^m,$$ 
and finally, for every $\color{blue}{c}$ and $\color{red}{a}$,

$$U_{m,n}(\color{red}{a},\color{red}{a}\color{blue}{c},\color{red}{a}\color{blue}{c}^2,\ldots,\color{red}{a}\color{blue}{c}^n)=(m\color{blue}{c})^n\color{red}{a}^m.$$

A: I proved my claim. I would like to share it.
Taylor expansion of $Z^m(x+h)$ can be written as
$$
Z^m(x+h)=Z^m(x)+h\frac{d}{dx} \left( Z^m(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( Z^m(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( Z^m(x) \right)+....
$$
If we derivate both sides n times over $h$ and put $h=0$
$$\frac{d^n}{dh^n} \left( Z^m(x+h) \right)|_{h=0}=\frac{d^n}{dx^n} \left( Z^m(x) \right)$$ 
Thus , we can write
$$ U(Z(x),Z'(x),Z''(x),Z'''(x),...,Z^{(n)}(x))=\frac{d^n}{dx^n} \left( Z^m(x) \right)=\frac{d^n}{dh^n} \left( Z(x)+hZ'(x)+h^2 \frac{Z''(x)}{2!}+h^3 \frac{Z'''(x)}{3!}+..... \right)^m |_{h=0}$$ 
$$ U(1,1,1,1,...,1)=\frac{d^n}{dh^n} \left( 1+h+h^2 \frac{1}{2!}+h^3 \frac{1}{3!}+..... \right)^m |_{h=0}=\frac{d^n}{dh^n} \left( e^{h} \right) ^m |_{h=0}=\frac{d^n}{dh^n} \left( e^{mh} \right) |_{h=0}=m^n \left( e^{mh} \right) |_{h=0}=m^n$$ 
