Let $T_1 \in \text{Exp}(\theta_1), \dots, T_N \in \text{Exp}(\theta_N)$ each be an D214: Exponential random positive real number such that

(i)	$T_1, \dots, T_N$ is an D2713: Independent random collection

Let $a \in \mathbb{R}$. If $a < 0$, then clearly $\mathbb{P} ( \min(T_1, \ldots, T_N) \leq a ) = 0$, so assume that $a \in [0, \infty)$. We have by independence \begin{equation} \begin{split} \mathbb{P} ( \min(T_1, \ldots, T_N) > a ) & = \mathbb{P} (T_1 > a, \ldots, T_N > a) \\ & = \prod_{n = 1}^N \mathbb{P}(T_n > a) \\ & = \prod_{n = 1}^N \int^{\infty}_a \theta_n e^{- \theta_n t} \, d t \\ & = \prod_{n = 1}^N \left[ - e^{\theta_n t} \right]^{\infty}_a = \prod_{n = 1}^N \left( 0 + e^{- \theta_n a} \right) = \prod_{n = 1}^N e^{- \theta_n a} = e^{- \left(\sum_{n = 1}^N \theta_n \right) a} \end{split} \end{equation} That is \begin{equation} \mathbb{P} ( \min(T_1, \ldots, T_N) \leq a ) = e^{- \left(\sum_{n = 1}^N \theta_n \right) a} \end{equation} The claim thus now follows from R4795: Real calculus expression for distribution function of exponential random positive real number. $\square$