Let $X$ be uniformly distributed over $(0,1)$. Let $Y=X^n$. Find the p.d.f. of $Y$. 
Let $X$ be uniformly distributed over $(0,1)$. Let $Y=X^n$. Find the
  p.d.f. of $Y$.

I tried using the transformation method:
$$h^{-1}(y)=y^{1/n}$$
$$\frac{dh^{-1}}{dy}=\frac{1}{n}y^{\frac{1}{n}-1}$$
$f_X(x)=x$ for $0<x<1$
$f_Y(y)=f_X(h^{-1})\frac{dh^{-1}}{dy} =\frac{1}{n}y^{\frac{2}{n}-1} $ for $0 < y < 1$.
Is this correct?
 A: The purpose of this Answer is to verify the answer for $n = 2$ and to
provide some intuition. In that case $Y = X^2 \sim \mathsf{Beta}(0.5, 1),$
with density function $f_Y(y) = 0.5y^{-0.5},$ for $0 < y < 1.$ The support
of $Y$ is $(0,1)$ because squaring values in $(0,1)$ give numbers in the
same interval.
An alternative derivation, using the so-called CDF method, is shown below.
This derivation also works for general $n.$
$$F_Y(y) = P(Y \le y) = P(X^2 \le y) = P(X \le \sqrt{y}) = \sqrt{y},$$
so density function is $f_Y(y) = F^\prime(y) = 0.5y^{-0.5},$ for $0 < y < 1.$
In the PDF method you used, students often wonder about an intuitive
explanation for the factor $dh^{-1}/dy.$ A random sampling experiment facilitates making
graphs that illustrate its role. The graphs below are made by sampling
100,000 observations $X \sim \mathsf{Unif}(0,1),$ squaring each one to
obtain 100,000 observations $Y \sim \mathsf{Beta}(0.5,1),$ and
making histograms that show what happens as a result of the transformation.
In the figure below, each histogram bar represents about 10,000 random
observations, and bars have equal areas. Points in a bar of a particular color in the histogram at the left are
transformed to a bar of the same color at the right. Narrower bars on the
right must be taller in order to have the same area; this illustrates 
the role played by $dh^{-1}/dy.$ Blue curves show the
density functions of the two distributions. (Ordinarily, making
histograms with bars of unequal widths is seldom a good idea, but it serves
a purpose here.)

Code to make the histograms in R statistical software is included below,
in case it is of any interest.
m = 10^5; x = runif(m); y = x^2
cut.x = seq(0, 1, by=.1); cut.y = cut.x^2
par(mfrow=c(1,2)); farb=rainbow(12)  # 'par' enables 2 histograms per plot
  hist(x, br=cut.x, prob=T, xlim=c(-0.1,1.1), ylim=c(0,10), col=farb, main="UNIF(0,1)")
    curve(dunif(x), col="blue", lwd=2, n=1001, add=T)
  hist(y, br=cut.y, prob=T, xlim=c(-0.1, 1.1), ylim=c(0,10), col=farb, main="BETA(.5,1)")
    curve(dbeta(x, .5, 1), col="blue", lwd=2, n=1001, add=T)
par(mfrow=c(1,1))  # return to single panel plots

