Suppose we have $N$ exponential random variables $X_i$ a with parameter $\beta$ $$P_{X_i}(x)=\beta e^{-\beta x}.$$ We define a random variable $Z_1$ which is the max of all $N$ exponential random variables $X_i$ as $$Z_1=\max\{X_1,\cdots,X_N\}.$$ The the PDF of $Z_1$ is $$P_{Z_1}(z)=N\beta e^{-\beta z}(1-e^{-\beta z})^{N-1}.$$ Using the binomial theorem, we will have at \begin{align} P_{Z_1}(z)=&N\beta\sum_{n=0}^{N-1}e^{-\beta z}\binom{N-1}{n}(-1)^n e^{-\beta zn} \\ = &N\beta\sum_{n=0}^{N-1}\binom{N-1}{n}(-1)^n e^{-\beta z(n+1)} \end{align} My question, can we simplify more than that.
What is the PDF of inverse of $Z_1$, $Y_1$ defined by $$Y_1=\frac{1}{Z_1}$$
Since $\mathbb P(Z_1>0)=1$, the distribution function of $Y_1$ is given by $$ F_{Y_1}(y) = \mathbb P(Y_1\leqslant y) = \mathbb P\left(Z_1\geqslant \frac1y\right) = 1 - F_{Z_1}\left(\frac1y\right), $$ where $F_{Z_1} = \left(1-e^{-\beta z}\right)^N$. Hence $$ F_{Y_1}(y) = 1 - \left( 1 - e^{-\frac\beta y}\right)^N, $$ and the density is obtained by differentiating: $$ f_{Y_1}(y) = \frac{\mathsf d}{\mathsf dy} \left[1 - \left( 1 - e^{-\frac\beta y}\right)^N\right] = \beta N e^{-\frac{\beta }{y}}y^{-2} \left(1-e^{-\frac{\beta }{y}}\right)^{N-1}. $$