We will assume that $N$ has right-continuous sample paths. Define the firth arrival time $T_1$ by
$$ T_1 = \inf \{ t \geq 0 : N_t \geq 1 \}. $$
By the right-continuity, this implies that $N_{T_1} = 1$. Then we will prove the following claim, which is a special case of the strong Markov property:
Claim. Define a new stochastic process $\tilde{N} = ( \tilde{N}_t )_{t\geq 0}$ by
$$ \tilde{N}_t = N_{T_1 + t} - N_{T_1} = N_{T_1 + t} - 1. $$
Then we have the following:
- $T_1$ and $\tilde{N}$ are independent.
- $\tilde{N}$ is a Poisson process of rate $\lambda$.
- $T_1$ is an exponential random variable of rate $\lambda$.
Proof. Fix $0 < t_0 < t_1 < \cdots < t_n$. Also, let $\delta > 0$ be sufficiently small so that $0 < t_0 - \delta$ and $t_{i-1}+\delta < t_i-\delta$ holds for all $i = 1, \dots, n$.
Now, for any parameters $s, s_1, \dots, s_n \geq 0$, we consider
$$ Y = sT_1 + \sum_{i=1}^{n} s_i (\tilde{N}_{t_i} - \tilde{N}_{t_{i-1}}). $$
Then, conditioned on $\{T_1 \in ((k-1)\delta, k\delta] \}$, we have
$$ Y \geq s (k-1)\delta + \sum_{i=1}^{n} s_i (N_{(k-1)\delta + t_i} - N_{k\delta + t_{i-1}}) $$
and so,
$$ \mathbb{E}[e^{-Y}]
\leq \sum_{k=1}^{\infty} \mathbb{E}\left[ \exp\left\{ - s (k-1)\delta - \sum_{i=1}^{n} s_i (N_{(k-1)\delta + t_i} - N_{k\delta + t_{i-1}}) \right\} \mathbf{1}_{\{T_1 \in ((k-1)\delta, k\delta] \}} \right]. \tag{1} $$
By noting that $\{T_1 \in ((k-1)\delta, k\delta] \} = \{ N_{(k-1)\delta} = 0 \text{ and } N_{k\delta} \geq 1 \} $, all of
$$ \mathbf{1}_{\{T_1 \in ((k-1)\delta, k\delta] \}}, \quad N_{(k-1)\delta + t_1} - N_{k\delta + t_{0}}, \quad \dots, \quad N_{(k-1)\delta + t_n} - N_{k\delta + t_{n-1}} $$
are independent, and hence the bound $\text{(1)}$ reduces to
\begin{align*}
\mathbb{E}[e^{-Y}]
&\leq \sum_{k=1}^{\infty} e^{-s(k-1)\delta} \left( \prod_{i=1}^{n} \mathbb{E}\left[ e^{-s_i (N_{(k-1)\delta + t_i} - N_{k\delta + t_{i-1}})} \right] \right) \mathbb{P} \left( T_1 \in ((k-1)\delta, k\delta] \right) \\
&= \sum_{k=1}^{\infty} e^{-s(k-1)\delta} \left( \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1} - \delta) (1 - e^{-s_i}) } \right) e^{-\lambda (k-1)\delta} (1 - e^{-\lambda \delta}) \\
&= \frac{1 - e^{-\lambda \delta}}{1 - e^{-(s+\lambda)\delta}} \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1} - \delta) (1 - e^{-s_i}) }. \tag{2}
\end{align*}
As $\delta \to 0^+$, the bound $\text{(2)}$ converges to
$$ \mathbb{E}[e^{-Y}] \leq \frac{\lambda}{s+\lambda} \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1}) (1 - e^{-s_i}) }. \tag{3} $$
By a similar computation applied to
$$ Y \leq s k\delta + \sum_{i=1}^{n} s_i (N_{k\delta + t_i} - N_{(k-1)\delta + t_{i-1}}), $$
we can also obtain the reverse direction of $\text{(3)}$. So it follows that
$$ \mathbb{E}[e^{-Y}] = \frac{\lambda}{s+\lambda} \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1}) (1 - e^{-s_i}) } \tag{4} $$
holds for any choices of parameters $s, s_1, \dots, s_n \geq 0$. However, since the multidimensional Laplace transform uniquely determines the law of a given random vector, this proves that:
$ T_1$, $\tilde{N}_{t_1} - \tilde{N}_{t_{0}}$, $\ldots$, $\tilde{N}_{t_n} - \tilde{N}_{t_{n-1}} $ are mutually independent,
$\tilde{N}_{t_i} - \tilde{N}_{t_{i-1}} \sim \operatorname{Poisson}(\lambda(t_i - t_{i-1}))$ for each $i = 1, \dots, n$, and
$T_1 \sim \operatorname{Exp}(\lambda)$.
Therefore the desired claim follows. $\square$
Now repeatedly applying this claim shows that inter-arrival times are independent exponential random variables of rate $\lambda$ as desired.
