This is taken from a stochastic processes text.
Consider $N$ a Poisson process (number of arrivals at time $t$) with rate $\lambda$. Let $T_n$ denote an arrival time ($n$ a natural number). The probability that there are no arrivals in $(t, t+s]$ is $e^{-\lambda s}$, independent of the history of arrivals before $t$. This can be applied to arrival times: $$P(N_{T_{n+s}}-N_{T_n} = 0 | N_u; u \leq T_n) = e^{-\lambda s}$$
The following proposition is provided:
For any $n$ a natural number,
$$P(T_{n+1}-T_n \leq t | T_0, ..., T_n) = 1 - e^{-\lambda t}, t \geq 0$$
My question is about the proposition. It appears that the probability is the complement of no arrivals. So why is the interval $\leq t$? I'm getting confused about the exponential expression appearing at first with number of arrivals whereas it is appearing again as a complement with times of arrivals.
You are right about the fact that the event $\{T_{n+1}-T_n \leq t\}$ is the complement of the event $\{N_{T_n+t}-N_{T_n}=0\}$.
The way to understand the proposition, therefore, is to consider the probability, $P(N_{T_n+t}-N_{T_n}=0|T_0,T_1,\ldots,T_n)$, which is nothing but $P(N_{T_n+t}-N_{T_n}=0|\{N_u:u\leq T_n\})$. This is because the collection of random variables $T_0,\ldots,T_n$ describes precisely the arrival process $\{N_u:u\leq T_n\}$.
Now, we know from the Strong Markov property that $P(N_{T_n+t}-N_{T_n}=0|\{N_u:u\leq T_n\})=e^{-\lambda t}$, like you've mentioned. The required probability, $P(T_{n+1}-T_n \leq t|\{N_u:u\leq T_n\}) = 1-e^{-\lambda t}$, by the complement property.