Knowing pdf of a random variable, make a transformation, how to find the moments after the transformation? The random variable X has the exponential probability density $f_X(x)=\alpha e^{-\alpha x}, x\geq0$

A new random variable is given by $Y=e^{-X}$. How can I obtain an expression for the n-th moment of Y.
The $n$-th moment of $Y=e^{-X}$ is $\langle Y^n\rangle=\langle e^{-nX}\rangle$. So you can just do it in the distribution of $X$ without having to transform the distribution to $Y$. The moment is

$$\langle Y^n\rangle=\int_0^\infty\alpha e^{-(n+\alpha)x}dx=\frac{\alpha}{\alpha+n}.$$

The distribution of $Y$ can also be found. Since $X\in[0,\infty)$, $Y=e^{-X}\in(0,1]$. The probability

$$P(Y&lt;y)=P(e^{-X}&lt;y)=P(X&gt;-\ln y)=y^\alpha,\quad y\in[0,1].$$

Then the density is $p(y)=\alpha y^{\alpha-1}$. From here one also obtains

$$\langle Y^n\rangle=\int_0^1\alpha y^{n+\alpha-1}dy=\frac{\alpha}{\alpha+n}.$$
The $n$-th moment of $Y=e^{-X}$ is $\langle Y^n\rangle=\langle e^{-nX}\rangle$. So you can just do it in the distribution of $X$ without having to transform the distribution to $Y$. The moment is

$$\langle Y^n\rangle=\int_0^\infty\alpha e^{-(n+\alpha)x}dx=\frac{\alpha}{\alpha+n}.$$

The distribution of $Y$ can also be found. Since $X\in[0,\infty)$, $Y=e^{-X}\in(0,1]$. The probability

$$P(Y&lt;y)=P(e^{-X}&lt;y)=P(X&gt;-\ln y)=y^\alpha,\quad y\in[0,1].$$

Then the density is $p(y)=\alpha y^{\alpha-1}$. From here one also obtains

$$\langle Y^n\rangle=\int_0^1\alpha y^{n+\alpha-1}dy=\frac{\alpha}{\alpha+n}.$$