Let $X \sim \exp(\lambda)$ and $\beta > 0$. I'm figure how to find $\mathbb{E}[X^\beta]$, that is
$$\mathbb{E}[X^\beta] = \lambda \int_0^{+\infty}t^\beta e^{-\lambda t} dt$$
I also know that an $\exp(\lambda)$ distribution can be seen as a $\Gamma(1, \lambda)$ distribution.
Since I know that for a $\Gamma(\alpha, \lambda)$ distribution it holds
$$\mathbb{E}[X^\beta] = \frac{\Gamma(\alpha + \beta)}{\lambda^\beta \Gamma(\alpha)}$$
therefore I would say that for $X \sim \exp(\lambda)$ I would obtain
$$\mathbb{E}[X^\beta] = \lambda \frac{\Gamma(\beta+1)}{\lambda^\beta} = \frac{\Gamma(\beta+1)}{\lambda^{\beta-1}}$$
But on textbook I have $\frac{\Gamma(\beta + 1)}{\lambda^\beta}$ as solution. What am I doing wrong?
You don't need to multiply by a factor of $\lambda$ again.
Since $$\mathbb{E}[X^\beta] = \lambda \int_0^{+\infty}t^\beta e^{-\lambda t} dt$$ and setting $\alpha=1$ we have $$\mathbb{E}[X^\beta] = \frac{\Gamma(\beta+1)}{\lambda^\beta\Gamma(1)} = \frac{\Gamma(\beta+1)}{\lambda^{\beta}}.$$
Alternatively let $x=\lambda t$ (with $\lambda>0)$ then we have $$\mathbb{E}[X^\beta] = \lambda \int_0^{+\infty}t^\beta e^{-\lambda t} dt$$ $$=\lambda\int_{0}^{\infty}\big(\frac{x}{\lambda}\big)^{\beta}e^{-x}\frac{dx}{\lambda}$$ $$=\frac{1}{\lambda^{\beta}}\int_{0}^{\infty}x^{\beta+1-1}e^{-x}dx$$ $$=\frac{\Gamma(\beta+1)}{\lambda^{\beta}}$$
where $\Gamma$ is the Gamma function.