As I mention here, the interarrival times $T_1, T_2, \ldots$ are distributed as $\text{Exponential}(\lambda)$ random variables. You seem to be interested in $T=\min (T_1,...,T_n)$ somehow, but as @gunes points out, you aren't clear about whether or not you're assuming that the total amount of time to get $n$ events must be less than $t$: "I have $n$ events in $t$ time." You need to tell us whether $t$ is random or not.
If it is, ot's more conventional to use uppercase letters, however capital $T$ is already taken to mean the minimum of all the interarrival times. The distribution of the minimum can easily be found using the following technique. Here is another link addressing the same issue.
If you are constraining the total amount of time, and conditioning on the fact that it's less than $t$, you are looking for the distribution of $T|\sum_{i=1}^n T_i < t$ as @gunes points out. The quickest way for me to find the cdf of this distribution is to use the Monte Carlo technique. To do this simulate $M$ sets of $n$ random variables $T_i^j$, and compute
\begin{align*}
P\left(T \le s \bigg\rvert \sum_{i=1}^n T_i < t\right) &= P\left(T \le s , \sum_{i=1}^n T_i < t\right)\bigg/ P\left(\sum_{i=1}^n T_i < t\right)\\
&\overset{\text{p}}{\leftarrow} \left(M^{-1}\sum_{j=1}^M 1\left\{T^j \le s , \sum_{i=1}^n T^j_i < t \right\}\right) \\
&\hspace{10mm} \bigg/ \left(M^{-1}\sum_{j=1}^M 1\left\{ \sum_{i=1}^n T^j_i < t .\right\}\right)\end{align*}
Here's some R code:
lambda <- 3
s <- .03
t <- 20
M <- 1000
n <- 20
samps <- matrix(rexp(M*n, lambda), nrow=M)
sum(apply(samps, 1, function(row) (min(row) < s) & (sum(row)< t) )) / sum(apply(samps, 1, function(row) (sum(row)< t)))
The Erlang distribution arises as the sum of iid exponential random variables. Instead of estimating $P\left(\sum_{i=1}^n T_i < t\right)$ with samples, you could swap in the known cdf function here, and it would reduce the variance of the above Monte Carlo estimator.
