The PDF of minimum of two random varible. Let $X$ and $Y$ be two random variables with respective PDFs $P_X(x)$ and $P_Y(y)$ given by
\begin{align}\label{Eq_Ap_2}
  P_{X}(x)&=M\beta e^{-\beta x}\begin{pmatrix}  1-e^{-\beta x}
\end{pmatrix}^{M-1}\\
        &=M\beta \sum_{m=0}^{N-1}(-1)^n\binom{M-1}{m}\begin{pmatrix}
                                                      e^{-\beta x}
                                                     \end{pmatrix}^{m+1}
\end{align}
\begin{align}
  P_{Y}(y)&=N\beta e^{-\beta y}\begin{pmatrix}  1-e^{-\beta y}
\end{pmatrix}^{N-1}\\
        &=N\beta \sum_{n=0}^{N-1}(-1)^n\binom{N-1}{n}\begin{pmatrix}
                                                      e^{-\beta y}
                                                     \end{pmatrix}^{n+1}.\label{Eq_Ap_3}
\end{align}
What is the PDF of 
$$
Z=\min\{X,Y\}
$$
 A: You did not state it, but I assume X, Y are independent
Firstly find the CDF of $Z = \min(X,Y)$ and then differentiate it w.r.t. z to get its PDF:
$\mathbb{P}(Z \leq z) = 1- \mathbb{P}(\min(X,Y) > z) = 1 - \mathbb{P}(X > z)\mathbb{P}(Y > z) = 1 - (1 - \mathbb{P}(X \leq z))(1 - \mathbb{P}(Y \leq z)) = \mathbb{P}(Y \leq z) + \mathbb{P}(X \leq z) - \mathbb{P}(Y \leq z)\mathbb{P}(X \leq z) $
Taking the derivative w.r.t. z yields:
$P_Z(z) = P_Y(z) + P_X(z) - P_Y(z)\mathbb{P}(X \leq z) - \mathbb{P}(Y \leq z)P_X(z) = P_X(z)\mathbb{P}(Y \geq z) + P_Y(z)\mathbb{P}(X \geq z) = P_X(z)\int_z^{\infty}P_Y(t)dt + P_Y(z)\int_z^{\infty}P_X(t)dt$
Substituting your PDFs and computing the integrals yields the final formula. Since the sum is finite you can interchange the integration and summation. Computing the integrals is not hard.
A: Assume X and Y are independent.  $P(Z\gt z)=P(X\gt z)P(Y\gt z)$. 
Thus $P(Z\lt z)=1-(1-P(X\lt z))(1-P(Y\lt z))=P(X\lt z)+P(Y\lt z)-P(X\lt z)P(Y\lt z).$
Therefore $P_Z(z)=P_X(z)+P_Y(z)-(CDF_X(z)P_Y(z)+CDF_Y(z)P_X(z))$. 
Note to comment - I was subconsciously using the formula for max rather than min before.    
A: The independence between $X$ and $Y$ will be assumed, otherwise there's no way to proceed.

TL;DR : see Eq.\eqref{Eq05} for the end result and an intuitive interpretation that follows. 

Note that $X$ can be viewed as the maximum of $M$ exponential random variables that are i.i.d. with rate of $\beta$. Denote this set as $U_i$ for $i = 1$ to $M$. All $U_i$ have the same density $f_U(t) = \beta\, \mathrm{Exp}(-\beta t)$. 
Similarly, $Y$ has the same distribution as the maximum of another set of $N$ exponential i.i.d with rate $\beta$. Call these $W_k$ for $k = 1$ to $N$, and their common density is the same $f_W(t) = \beta\, \mathrm{Exp}(-\beta t)$. 
The two sets $U_i$ and $W_k$ are independent, otherwise $X$ and $Y$ cannot be independent.
Denote the cumulative function as $F_X(t) \equiv \Pr\{ X < t\}$, not to be confused with the density $P_X(x)$. The same notation goes for $F_U(t)$, $F_W(t)$, and $F_Y(t)$. We are given
\begin{align*}
F_X(t) &= \bigl( F_U(t) \bigr)^M = \left( 1 - e^{-\beta t} \right)^M  & &\text{and} & F_Y(t) &= \bigl( F_W(t) \bigr)^N = \left( 1 - e^{-\beta t} \right)^N
\end{align*}
There is actually an intuitive shortcut to obtain the density of $Z = \min\{X, Y\}$, as is often the case with order statistics. Nonetheless, let's do it concretely from the cumulative function to appreciate some inner structure.

\begin{align*}
F_Z(t) \equiv \Pr\{ Z < t\} &= \Pr\{ X~\text{is smaller} \} + \Pr\{ Y~\text{is smaller} \} \\
&= \color{blue}{ \Pr\{ X < t ~~\&~~ Y> X \}} + \Pr\{ Y < t ~~\&~~ X > Y \} \\
&= \color{blue}{ \int_{x = 0}^t \int_{y=x}^{\infty} P_X(x) P_Y(y)\,\mathrm{d}y \,\mathrm{d}x } + \int_{y = 0}^t \int_{x=y}^{\infty} P_X(x) P_Y(y)\,\mathrm{d}x \,\mathrm{d}y \tag{1} \label{Eq01}
\end{align*}
In the above the fact is already invoked that $X \perp Y \implies P_{XY}(x,y) = P_X(x)P_Y(y)$ 
Consider the first integral in Eq.\eqref{Eq01}, which is the region above the diagonal in the $X$-$Y$ plane.
\begin{align*}
\color{blue}{ \Pr\{ X < t ~~\&~~ Y> X \} } &= \int_{x = 0}^t \left[ P_X(x)\int_{y=x}^{\infty} P_Y(y)\,\mathrm{d}y \right] \,\mathrm{d}x \\ 
&= \int_{x = 0}^t \Bigl[ P_X(x) \bigl( 1 - F_Y(x) \bigr)\Bigr] \,\mathrm{d}x \\
&= \int_{x = 0}^t P_X(x) \cdot 1 \,\mathrm{d}x - \int_{x = 0}^t  P_X(x) \cdot F_Y(x) \,\mathrm{d}x \\
&= F_X(t) - \int_{x = 0}^t  M \beta e^{-\beta x} \left( 1 - e^{-\beta x} \right)^M \cdot \left( 1 - e^{-\beta x} \right)^N \,\mathrm{d}x 
\end{align*}
Since in the end we will take the derivative with respect to $t$ to obtain the density as in
$$P_Z(t) = \frac{ \mathrm{d}F_Z(t) }{ \mathrm{d} t} = \color{blue}{ \frac{ \mathrm{d}\Pr\{ X < t ~~\&~~ Y> X \} }{ \mathrm{d} t} } + \frac{ \mathrm{d}\Pr\{ Y < t ~~\&~~ X > Y \} }{ \mathrm{d} t} \tag{2} \label{Eq02}$$
we might as well do it now for the above-diagonal piece.
$$\color{blue}{ \frac{ \mathrm{d}\Pr\{ X < t ~~\&~~ Y> X \} }{ \mathrm{d} t}}  = P_X(t) 
 - \color{magenta}{\frac{ M }{ M+N } } (M+N) \beta e^{-\beta t} \left( 1 - e^{-\beta t} \right)^{M+N} \tag{3} \label{Eq03}$$
After the leading $P_X(t)$ we get the $\color{magenta}{\text{scaled}}$ version of the density of the maximum of $(M+N)$ exponential i.i.d. with rate $\beta$.
Similarly, that the second piece of integral in Eq.\eqref{Eq01} for the below-diagonal region will give:
$$\frac{ \mathrm{d}\Pr\{ Y < t ~~\&~~ X> Y \} }{ \mathrm{d} t} = P_Y(t) 
 - \color{magenta}{\frac{ N }{ M+N } } (M+N) \beta e^{-\beta t} \left( 1 - e^{-\beta t} \right)^{M+N} \tag{4} \label{Eq04}$$
Put Eq.\eqref{Eq03} and Eq.\eqref{Eq04} them together according to Eq.\eqref{Eq02} we get a full piece.
$$P_Z(t) = P_X(t) + P_Y(t) - P_T(t) \quad \text{where} \quad P_T(t) = (M+N) \beta e^{-\beta t} \left( 1 - e^{-\beta t} \right)^{M+N} \tag{5} \label{Eq05}$$ 

That is, the density of $Z = \min\{X,Y\}$ is the sum of the densities of $X$ and $Y$ minus the density of $T$, where $T$ is the maximum of $(M+N)$ exponential i.i.d. with rate $\beta$

There's actually a heuristic argument for this "sum then minus" of densities. Note that here we have only two things $X$ and $Y$, therefore
$$X + Y = \min\{X, Y\} + \max\{X, Y\} $$
Many nice properties come out of this fact, and there are some nice example for the discrete distribution as well. In particular, consider the infinitesimal probability mass between (the dummy) $t$ to $t+ \mathrm{d}t$. With notations $Z = \min\{X, Y\}$ and $T = \max\{X, Y\}$, we have 
$$\Pr\{ t < X < t+ \mathrm{d}t \} + \Pr\{ t < Y < t+ \mathrm{d}t \} = \Pr\{ t < Z < t+ \mathrm{d}t \} + \Pr\{ t < T < t+ \mathrm{d}t \} \tag{6} \label{Eq06}$$
Intuitively, the tiny probability mass is proportional to the density as in 
$$\Pr\{ t < X < t+ \mathrm{d}t \} = P_X(t)\mathrm{d}t \tag*{, similar for $Y$, $Z$, and $T$.}$$
Therefore, Eq.\eqref{Eq06} becomes an equation of the densities.
$$P_X(t) + P_Y(t) = P_Z(t) + P_T(t) \\
P_Z(t) = P_X(t) + P_Y(t) - P_T(t)$$
This can be considered an explanation for Eq.\eqref{Eq05} as well as an intuitive shortcut to derive it.
