showing a random variable has an exponential distribution Let $X_{1},..,X_{n}$ be independent, each with a exp($\lambda$) distribution. Let $Z=min(X_{1},..X_n)$. Show that $n\lambda Z$ has an exp$(1)$ distribution. 
I calculate that $P(Z>z)=e^{-n\lambda z}$. Hence the density function of $n\lambda Z$ is $f(z)=(n\lambda)^2e^{-n\lambda z}$. But an $exp(1)$ distribution would have density $e^{-z}$. 
 A: The distribution function (CDF) of the minimum order statistic from an i.i.d. sample of $X$'s is
$$F_Z(z) = 1- [1- F_X(z)]^n$$
In our case $F_X(x) = 1-e^{-\lambda x}$ so
$$F_Z(z) = 1- [1- (1-e^{-\lambda z})]^n = 1-e^{-n\lambda z}$$
We are interested in the distribution of the random variable $Y\equiv n\lambda Z$.
A) The "CDF method" goes as follows:
$$P(Y \leq y) = P(n\lambda Z \leq y) = P\left( Z \leq \frac 1{n\lambda}y\right) = F_Z\left(  \frac 1{n\lambda}y\right) = 1-e^{-n\lambda \frac 1{n\lambda}y}$$
$$\implies F_Y(y) = 1-e^{-y}$$
which is the CDF of an $\text{Exp}(1)$ random variable. In general, when the cumulative distribution function of the "source" random variable is closed formed/simple, this method is less prone to mistakes than the
B) "Change of Variable" method  
The density function of $Z$ is
$$ f_Z(z) = n\lambda e^{-n\lambda z}$$
For the random variable $Y\equiv n\lambda Z$ we have $Z = \frac 1{n\lambda}Y$ and so 
$$f_Y(y) = \left|\frac {\partial Z}{\partial Y}\right|\cdot f_Z\left(\frac 1{n\lambda}y\right) = \frac 1{n\lambda}\cdot n\lambda e^{-n\lambda \frac 1{n\lambda}y} = e^{-y}$$
which is the probability density function of an $\text{Exp}(1)$ random variable.
