Proof
Using the Chi-Square probability function 9.3.1,
\begin{align*}
M(t) & = \int_0^{\infty} e^{tx} \frac{x^{r/2-1} e^{-x/2} }{\Gamma(r/2) 2^{r/2} } dx\\
& = \int_0^{\infty} \frac{x^{r/2-1} e^{-x(1-2 t)/2} }{\Gamma(r/2) 2^{r/2}} dx\\
& = \int_0^{\infty} \frac{\left ( \frac{u}{1-2t} \right )^{r/2-1} e^{-u/2} }{\Gamma(r/2) 2^{r/2} } \frac{1}{1-2t} du\\
& = \left ( \frac{1}{1-2t} \right )^{r/2} \int_0^{\infty} \frac{u^{r/2-1} e^{-u/2} }{\Gamma(r/2) 2^{r/2}} du\\
& = \left ( \frac{1}{1-2t} \right )^{r/2}
\end{align*}
where the final integral is again Chi-Square and therefore has value 1.