Given $Y_i $ with pdf $f(y) = my^{m-1}/n^m, m>0, n>0, 0<y\leq n, 1\leq i \leq k$, prove that $-k\ln (\frac{Y_{(k)}}{n})$ is Gamma$(1,1/m)$.
I tried finding the CDF of $f$. I used that to find the CDF of the largest order statistic using the formula $P(Y_{(k)}<x) = \prod^k_iP(Y_i<x)$. Then, I used a one to one transformation to find $-k\ln (Y_{(k)}/n)$ but it didn't get me the right distribution.
Am I tackling this the right way?
The first thing to observe is that $Y_i$ is a scaled transformation of a beta distribution; specifically, $Y_i = n X_i$, where $$X_i \sim \operatorname{Beta}(m, 1), \\ f_{X_i}(x) = mx^{m-1} \mathbb 1(0 < x < 1).$$ Therefore, $Y_{(k)} = n X_{(k)}$, and $$F_{X(k)}(x) = \Pr[X_{(k)} \le x] = \prod_{i=1}^k \Pr[X_i \le k] = \left(\int_{t=0}^x mt^{m-1} \, dt\right)^k = (x^m)^k = x^{mk}.$$ Thus $$\frac{Y_{(k)}}{n} = X_{(k)} \sim \operatorname{Beta}(mk, 1).$$ It follows that $$W = - k \log \frac{Y_{(k)}}{n}$$ has density $$f_W(w) = f_{X_{(k)}}(e^{-w/k}) \left| \frac{d}{dw}\left[ e^{-w/k} \right]\right| = mk(e^{-w/k})^{mk-1} \cdot \frac{e^{-w/k}}{k} = m e^{-mw} \mathbb 1(w > 0),$$ that is to say, $$W \sim \operatorname{Exponential}(\lambda = m).$$