How can we prove $\sum\limits_{k=0}^{\infty}(m+k+1)^nx^k=\sum\limits_{q=0}^{n}\binom{n}{q}m^{n-q}\frac{A_{q}(x)}{(1-x)^{q+1}}$?

Question

If $$\sum\limits_{k=0}^{\infty}(k+1)^nx^k=\frac{A_{n}(x)}{(1-x)^{n+1}}$$ so $$\sum\limits_{k=0}^{\infty}(m+k+1)^nx^k=\sum\limits_{q=0}^{n}\binom{n}{q}m^{n-q}\frac{A_{q}(x)}{(1-x)^{q+1}}$$ How can we prove it?

Answer 1

\begin{align} \sum_{k=0}^\infty (m+(k+1))^n x^k &= \sum_{k=0}^\infty \sum_{q=0}^n \binom{n}{q}(k+1)^q m^{n-q} x^k\\ &= \sum_{q=0}^n \binom{n}{q}m^{n-q}\sum_{k=0}^\infty (k+1)^q x^k \\ &= \sum_{q=0}^n \binom{n}{q}m^{n-q}\frac{A_q(x)}{(1-x)^{q+1}} \\ \end{align}

Answer 2

Hint. One may observe that, by the binomial theorem, $$ (m+k+1)^n=\sum\limits_{q=0}^{n}\binom{n}{q}m^{n-q}(k+1)^q. $$