I'm studying basic probability and in my lecture notes, it shows how the Gamma function results from the convolution of two exponential random variables. To introduce the gamma function it shows that for a random variable $X \sim \mathrm{Gamma}(n,\lambda)$:
$$ \int_{-\infty}^\infty f(x) \ dx = \int_{- \infty}^\infty \frac{\lambda^n x^{n-1} e^{-\lambda x}}{(n-1)!} \ dx = \int_{- \infty}^\infty \frac{\lambda (\lambda x)^{n-1}e^{- \lambda x}}{(n-1)!} \ dx = 1$$
and so by setting $ z = \lambda x $...
$$ (n-1)! = \int_0^\infty z^{n-1} e^{-z} \ dz$$
The step $ \int_{- \infty}^{\infty} \frac{\lambda(\lambda x)^{n-1}e^{- \lambda x}}{(n-1)!} \ dx = 1$ is left unexplained and I am confused as to how my lecturer is obtaining this result.
If they are taking for granted that the Gamma mass function is well-defined and that the unit measure axiom is satisfied , how can I prove this for myself and that this reasoning makes sense? Otherwise, what has my lecturer done in this step? Is there a way to easily integrate this function?