We know that the exponential distribution characterises the probability distribution for the waiting time between two consecutive Poisson events. Then I think if we fix a time interval $T$ then we should be able to derive the Poisson distribution from the exponential distribution. But some problems arises during my derivation.
Assuming the number of events per unit time is $\lambda$, then we have the Poisson distribution for the time interval $[0,T]$ $$P(k)=\frac{(\lambda T)^ke^{-\lambda T}}{k!}$$ Now we start with the exponential distribution $f(t)=\lambda e^{-\lambda t}$, for $t\ge 0$, and try to calculate, for instance, $P(k=0)$, $P(k=1)$ and $P(k=2)$, to see whether it gives the same result with the general Poisson distribution.
For $k=0$, no events happen in the time interval. Therefore, $$P(k=0)=\int_T^{\infty}f(t)\mathrm{d}t=e^{-\lambda T},$$ which agrees with the standard Poisson form.
For $k=1$, we have to make sure that one event happens at time $t_1 \in [0,T]$ and the the second event happens after $T$, i.e. $0\le t_1\le T \le t_2...$. Therefore, we have $$P(k=1)=\int_0^T\mathrm{d}t_1f(t_1)\int_{T-t_1}^{\infty}\mathrm{d}t'f(t')=\lambda Te^{-\lambda T}$$ where the second integral simply means the waiting time $t''$ between the two events has to be larger than $T-t_1$ to guarantee that the second event happens outside the interval $[0,T]$. Again, this result agrees with the standard Poisson distribution.
For $k=2$, we then have $0\le t_1 \le t_2 \le T \le t_3...$ and similarly we have the following $$P(k=2)=\int_0^T\mathrm{d}t_1f(t_1)\int_{0}^{T-t_1}\mathrm{d}t'f(t')\int_{T-(t_1+t')}^{\infty}\mathrm{d}t''f(t'')=\frac{(\lambda T)^2e^{-\lambda T}}{2}$$ Edited version, thanks to Kevin Zhou. Once the limited is corrected, a correct result will be obtained.
