I found this question
Prove that for all $n\in \Bbb N_0$ there exists a unique polynomial $Q_p\in \Bbb Q[x]$ such that $$\left\{\begin{array}{l}Q_p(x+1)-Q_p(x)=x^p\quad (\star)\\ Q_p(0)=0\end{array}\right.$$ and that we have $$Q'_p(x)-Q'_p(0)=pQ_{p-1}(x),\quad \forall p\in \Bbb N$$
Well, for the first part of the question, I defined the linear transformation $\varphi$ on the linear space $\Bbb Q[x]$ defined by: $\varphi(Q)=Q(x+1)-Q(x)$. I proved that $\ker\varphi=\Bbb Q_0[x]$ i.e. the constant polynomials and that $\operatorname{Im}\varphi=\Bbb Q[x]$ so that for $x^p$ there is $Q_p$ such that $\varphi(Q_p+c)=x^p$ for all constant $c\in \Bbb Q$. Now we choose $c=-Q_p(0)$ to satisfy the requirement $Q_p(0)=0$ hence the uniqueness. Now I'm stuck in the second part of the question. I tried to differentiate the equality $(\star)$ and to replace $px^{p-1}$ by $p(Q_{p-1}(x+1)-Q_{p-1}(x))$ but I can't progress further. Any idea?
First part is ok.
For the second one, differentiate as you did : you have $$Q_p'(x) = Q_p'(x+1)-px^{p-1} = Q_p'(x+1)-p(Q_{p-1}(x+1)-Q_{p-1}(x))$$ $$= Q_p'(x+1) - pQ_{p-1}(x+1) + pQ_{p-1}(x) $$
So you get that for each $x$, $$Q_p'(x) - pQ_{p-1}(x)= Q_p'(x+1) - pQ_{p-1}(x+1) $$
This shows that the polynomial $Q'_p - pQ_{p-1}$ is constant. Evaluating in $0$ shows that its value is equal to $Q_p'(0)$.