How to result in moment generating function of Weibull distribution? So I'm trying to get the Weibull distribution moment generating function 
$$\sum_{n=0}^\infty \frac{t^n \lambda^n}{n!} \Gamma(1+n/k)$$
(which can be found here https://en.wikipedia.org/wiki/Weibull_distribution)
I'm trying to do it with the definition of moment generating functions, but I can't get to that result. If $X$ has $\mathrm{Weibull}(\lambda, k)$ distribution, then its PDF is
$$f(x) =
\begin{cases}
\frac{k}{\lambda}(\frac{x}{\lambda})^{k-1} e^{(-x/\lambda)^k},  & \text{if $x\ge0$} \\
0, & \text{if $x<0$}
\end{cases} $$
And its moment generating function should be
$$M(t)=E(e^{tX})=\int_0^\infty e^{tx}\frac{k}{\lambda} \left(\frac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k} dx$$
but I can't get this form from Wikipedia. I have tried many different variable changes, etc., but without good result.
I appreciate any help leading to that form.

$$\begin{align}
M(t) &= E(e^{tX}) = \int_0^\infty
e^{tx}\frac{k}{\lambda} \left(\frac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k} dx
&\\ &= \int_0^\infty e^{\lambda tu} ku^{k-1} e^{-u^k} du
&\qquad(\text{with }u=x/\lambda\text{, for }\lambda>0)
\\ &= \int_0^\infty e^{\lambda tx^{1/k}} e^{-x} dx
&\qquad(\text{with }x=u^k\text{, for }k>0)
\\ &= \int_0^\infty \sum_{n=0}^\infty
 {(\lambda t)^n\over n!} x^{n/k} e^{-x} dx
\\ &= \sum_{n=0}^\infty {(\lambda t)^n\over n!}
 \int_0^\infty x^{n/k} e^{-x} dx
\\ &= \sum_{n=0}^\infty {\lambda^n t^n \over n!}
 \Gamma\left({n\over k}+1\right)
\end{align}$$
from definitions with two substitutions (as indicated),
the series expansion (for $e^{\lambda tx^{1/k}}$),
and integrating term by term (with the definition of the
Gamma function).
