It isn't proved in my study book of bio-statistics but suggested as an exercise. I tried the following method:
$PMF(y) = a*((PMF(x))^b$ = $a*((k/θ)((x/θ)^(k-1))e^(-(x/θ)^k))^b$
= $a*(k/θ)^b((x/θ)^(kb-b))(e^(-(x/θ)^(kb)))$
= $a*(k^b/θ^b)(x^k/θ^k)(e^(-x^(k*b)/θ^(k*b)))$
if k = s-1
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-x^(bl-b)/θ^(bl-b)))$ = $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-(x/θ)^s))$
So if I didn't make a mistake, I become something that looks like a Weibull distribution, but $a(s-1^b/θ^b)$ should be s-1/θ. I have no idea how I can become that if a,b is random and a,b>0
I searched on the web, but I couldn't find something similar. I hope someone can help me or suggest an other approach.
Other approach:
Let $U$ have standard exponential distribution.
Then for $x>0$: $$P\left(\theta U^{\frac{1}{k}}>x\right)=P\left(U\geq\left(\frac{x}{\theta}\right)^{k}\right)=e^{-\left(\frac{x}{\theta}\right)^{k}}$$ showing that $\theta U^{\frac{1}{k}}$ has Weibull distribution with parameters $\theta$ and $k$.
Conversely if $W$ has Weibull distribution then $U:=\left(\frac{W}{\theta}\right)^{k}$ has standard exponential distribution and we can write $W=\theta U^{\frac{1}{k}}$.
Then $Y=aX^{b}=a\left(\theta U^{\frac{1}{k}}\right)^{b}=a\theta^{b}U^{\frac{1}{kb^{-1}}}$ showing directly that $Y$ has Weibull distribution with parameters $a\theta^{b}$ and $kb^{-1}$.