Let $X\in NB(r,p)$and $Y\in NB(r+1,p)$, where $0<p<1$ and $r\ge 1$. Show $\Bbb E[X^n]=\frac {r}{p}\Bbb E[(Y-1)^{n-1}], n\ge 1$.
I feel that $\Bbb E(X^n)$ is the same thing with the mgf of negative binomial distribution, but I don't know how exactly can I manipulate the mgf to make it look like the right hand side.
Recall that for a negative binomial random variable $X$ with parameters $r$ and $p$, we have PMF $$\Pr[X = x] = \binom{x-1}{r-1} p^r (1-p)^{x-r}, \quad x \in \{r, r+1, \ldots \}.$$ Thus $$\begin{align*} \operatorname{E}[X^n] &= \sum_{x=r}^\infty x^n \binom{x-1}{r-1} p^r (1-p)^{x-r} \\ &= \sum_{x=r}^\infty x^{n-1} \frac{x!}{(r-1)! (x-r)!} p^r (1-p)^{x-r} \\ &= \frac{r}{p} \sum_{x=r}^\infty x^{n-1} \frac{((x+1)-1)!}{((r+1)-1)! ((x+1)-(r+1))!} p^{r+1} (1-p)^{(x+1)-(r+1)} \\ &= \frac{r}{p} \sum_{x=r}^\infty ((x+1) - 1)^{n-1} \binom{(x+1)-1}{(r+1)-1} p^{r+1} (1-p)^{(x+1)-(r+1)} \\ &= \frac{r}{p} \sum_{y=r+1}^\infty (y-1)^{n-1} \binom{y-1}{(r+1)-1} p^{r+1} (1-p)^{y-(r+1)} \\ &= \frac{r}{p} \operatorname{E}[(Y-1)^{n-1}]. \end{align*}$$