$E(Y^4)$ for exponential distribution. I was doing a question on finding the variance for an exponential distribution and happened on the term $E(Y^4)$. We have the pdf of the exponential distribution as $f(y) = \frac{1}{\beta}e^{\frac{-y}{\beta}}$ where we have $\beta = 10$. So, then we have $$E(Y^4) = \int_0^\infty y^4\frac{1}{10}e^{\frac{-y}{10}} \, dy$$ I was thinking that this looked awfully familiar to the $\Gamma$ function so I set $$E(Y^4) = 10^3\Gamma(5)$$ where $y = \frac{y}{10}$ and $$\Gamma(a) = \int_0^\infty y^{a-1}e^{-y} \, dy$$ But the answer I'm given in the solution manual is $E(Y^4) = 10^4\Gamma(5)$. Could someone please explain why they got the result they did? Why is mine wrong? Thanks.
Note that
$$E(Y^4) = \int_0^\infty y^4e^{\frac{-y}{10}}\frac{1}{10}\,dy
=\int_0^\infty (10u)^4e^{-u}\,du
=10^4\int_0^\infty u^4e^{-u}\,du
=10^4\Gamma(5)
$$
where we made the substituion $u=y/10$.
$$
\operatorname{E}(Y^4) = \int_0^\infty y^4\frac{1}{10} e^{-y/10} \, dy = 10^4 \int_0^\infty \left( \frac y {10} \right)^4 e^{-y/10}\left( \frac{dy}{10} \right) = 10^4 \int_0^\infty u^4 e^{-u} \, du.
$$