Let $X_1, X_2, X_3, \ldots \in \text{Bernoulli}(\theta)$ be an independent collection of Bernoulli random numbers such that
\begin{equation}
N
\overset{d}{=} \min \left\{ M \in \{ 1, 2, 3, \ldots \} : \sum_{m = 1}^M X_m = 1 \right\}
\end{equation}
The case $n = 0$ clearly holds since $\mathbb{P}(N > 0) = 1 = (1 - \theta)^0$. Suppose thus that $n \geq 1$. We have
\begin{equation}
\{ N > n \}
= \left\{ \sum_{m = 1}^n X_m = 0 \right\}
= \{ X_1 = 0, X_2 = 0, \ldots, X_n = 0 \}
\end{equation}
Thus by independence
\begin{equation}
\mathbb{P}(N > n)
= \mathbb{P} \left( X_1 = 0, X_2 = 0, \ldots, X_n = 0 \right)
= \prod_{m = 1}^n \mathbb{P}(X_m = 0)
= (1 - \theta)^n
\end{equation}
$\square$
