Finding integral of $2 \lambda\int_0^{\infty} x^{2n}xe^{-\lambda x^2} \ dx$ Finding integral of $$2 \lambda\int_0^{\infty} x^{2n}xe^{-\lambda x^2} \ dx$$
using integral by parts I get that Finding integral of $$2 \lambda\int_0^{\infty} x^{2n}xe^{-\lambda x^2} \ dx = \dfrac{n}{\lambda} \cdot \dfrac{n-1}{\lambda} \int_0^\infty x^{2n-4} \cdot x e^{-\lambda x^2} \ dx$$
so I assume that $$2 \lambda\int_0^{\infty} x^{2n}xe^{-\lambda x^2} \ dx = \dfrac{n!}{\lambda^n}$$. How can I formally prove this?

$$2 \lambda\int_0^{\infty} x^{2n}xe^{-\lambda x^2} \ dx =\frac{1}{\lambda^n}\int_0^{\infty} (\lambda x^2)^{n}e^{-\lambda x^2} \ d(\lambda x^2)\\\overset{\color{blue}{t= \lambda x^2}}{=}\frac{1}{\lambda^n}\int_0^{\infty} t^{n}e^{-t} \ dt =\frac{\Gamma(n+1)}{\lambda^n} =\frac{n!}{\lambda^n}$$
