Negative Binomial mean Let $X\sim BN(r,p)$.
If I try to compute E[X] through the Moment Generating Function I get the following:
\begin{aligned}
E[X] &=\left.\frac{d}{d t} M_{X}(t)\right|_{t=0} \
&=\left.\frac{d}{d t} p^{r}\left(1-p e^{t}(1-p)\right)^{-r}\right|_{t=0} \
&=p^{r+1} r(1-p)(1-p(1-p))^{-r-1}
\end{aligned}
But if I find it through the first factorial moment I get
$E\left[X^{(h)}\right]$
$=\sum_{x=0}^{\infty} \frac{x !}{(x-h) !} \frac{(x+r-1) !}{x !(r-1) !} p^{r}(1-p)^{x}$
$=\sum_{x=h}^{\infty} \frac{(x+r-1) !}{(x-h) !(r-1) !} p^{r}(1-p)^{x},\left(x^{*}=x-h\right)$
$=\sum_{x^{*}=0}^{\infty} \frac{\left(x^{*}+r+h-1\right) !}{x^{*} !(r-1) !} p^{r}(1-p)^{x^{*}+h}$
$=\frac{(r+h-1) !}{(r-1) !} \frac{(1-p)^{h}}{p^{h}} \sum_{x^{*}=0}^{\infty} \frac{\left(x^{*}+h+r+1\right) !}{x^{*} !(r+h-1) !} p^{r+h}(1-p)^{x^{*}}$
$=\frac{(r+h-1) !}{(r-1) !} \frac{(1-p)^{h}}{p^{h}}$

Therefore,

$E[X]=\frac{r(1-p)}{p}$

Where did I do wrong?

Your second derivation seems right. I can only guess that you made some minor error taking derivative.
Maybe it is better to compute abstract: Let $M_Y(t)$ be the moment generating function of a geometrically distributed random variable $Y$ with parameter $p\in(0,1)$, then $M_X(t)=(M_Y(t))^r$.
Therefore by chain rule
$$M_X(t)'=r(M_Y(t))^{r-1}\cdot M_Y'(t)$$
Taking $t$ to zero, the result is just $\mathbb E[X]=r\cdot\mathbb E[Y]=r\cdot\frac{1-p}{p}$, as you correctly derived using the second method.
