Q: Relation between Weibull and exponential distributions The probability distribution function of a Weibull distribution is as follows: 
$$
f(x) = a\cdot b^{-a}x^{a-1}\cdot e^{(-x/b)^a},\quad x>0
$$
for parameters $a,b>0$.
I have to show that $X\sim\mathrm{Weibull}(a,b)$ iff $X^a\sim\mathrm{expo}(b^a)$. Please help me to solve this question. This problem is taken from excercise of "Simulation Modeling And Analysis" book. If there is an solution book to you, I will be greatly helpful you can give that to me.  

A: Hint: Argue in terms of the cumulative distribution function (CDF). 
If $X\sim \mathrm{Weibull}(a,b)$, then $X$ has CDF $F_X$ given by
$$
F_X(x)=P(X\leq x)=
\begin{cases}
1-\exp\left(-(x/b)^a\right),\quad &x>0,\\
0,&x\leq 0.
\end{cases}
$$
Similarly, if $Y\sim \mathrm{exp}(\lambda)$, then $Y$ has CDF $F_Y$ given by
$$
F_Y(y)=
\begin{cases}
1-\exp(-\lambda x),\quad &y>0,\\
0,&y\leq 0.
\end{cases}
$$
So for the first direction, assume that $X\sim \mathrm{Weibull}(a,b)$ for some $a,b>0$. Then we aim at finding the CDF for $X^a$ and confirm that this corresponds to the CDF of an exponential distribution with parameter $b^a$. For $x\leq 0$ we have $P(X^a\leq x)=0$ and for $x>0$ we have
$$
P(X^a\leq x)=P(X\leq x^{1/a})=\ldots
$$
and hence...
