The question is not clear, but maybe you'll find the following useful.
For $t > 0$ fixed, let $E_k$, for $k=1,\ldots,n$, denote the event of $0$ events in the time interval $I_k = [\frac{{(k - 1)t}}{n},\frac{{kt}}{n}]$. Since $I_k$ has length $t/n$, $ {\rm P}(E_k^c) = \lambda \frac{{ t}}{n} + o\bigg(\frac{1}{n}\bigg) \;\; {\rm as} \;\; n \to \infty. $ Hence $ {\rm P}(E_k) = 1 - \lambda \frac{{ t}}{n} + o\bigg(\frac{1}{n}\bigg) \;\; {\rm as} \;\; n \to \infty. $ Now, by independence, the probability $p_t$ of $0$ events in the time interval $[0,t]$ is given by $ p_t = {\rm P}(E_1 ) \cdots {\rm P}(E_n ) = \bigg(1 - \frac{{\lambda t}}{n} + o\bigg(\frac{1}{n}\bigg)\bigg)^n . $ Letting $n \to \infty$, this gives $ p_t = e^{ - \lambda t} . $ This probability corresponds to ${\rm P}(X > t)$, where $X$ is exponential with parameter $\lambda$.
EDIT: Note that $n o(1/n) \to 0$ as $n \to \infty$. Using this, it indeed follows that $ \bigg(1 - \frac{{\lambda t}}{n} + o\bigg(\frac{1}{n}\bigg)\bigg)^n \to e^{ - \lambda t} . $