Let $(\sigma_i)$ be an i.i.d. sequence of exponentially distributed waiting times:
$$
\mathbb P(\sigma_i\in[t,t+dt])=\lambda\,e^{-\lambda t}\,dt
$$
and define $\tau_n:=\sum_{i=1}^n\sigma_i$ to be the time of the $n$-th jump. The jump process is defined as
$$
N_t=\sum_{n=1}^\infty 1_{\{\tau_n\le t\}}\,.
$$
The distribution of $N_t$ is obtained from
\begin{align}
&\mathbb P(N_t= k)\\&=\mathbb P\left(\tau_k\le t<\tau_{k+1} \right)=\mathbb P\left(\sigma_1+...+\sigma_k\le t<\sigma_1+...+\sigma_{k+1} \right)\\
&=\mathbb P\Big(\sigma_1\le t,\sigma_2\le t-\sigma_1,...,\sigma_k\le t-\sigma_1-...-\sigma_{k-1},t-\sigma_1-...-\sigma_{k} <\sigma_{k+1} \Big)\\ \tag{1}
&=\int_0^t\int_0^{t-s_1}...\int_0^{t-s_1-...-s_{k-1}}\int_{t-s_1-...-s_k}^\infty\lambda^{k+1}e^{-\lambda s_1-...-\lambda s_{k+1}}\,ds_{k+1}\,...ds_1\,.
\end{align}
The $ds_{k+1}$-integral is
\begin{align}
\lambda^ke^{-\lambda s_1-...-\lambda s_k}\,e^{-\lambda t+\lambda s_1+...+\lambda s_k}
=\lambda^ke^{-\lambda t}\,.
\end{align}
The $ds_k$-integral is
\begin{align}
\lambda^k e^{-\lambda t}\big(t-s_1-...-s_{k-1}\big)\,.
\end{align}
The $ds_{k-1}$-integral is
$$
\lambda^k e^{-\lambda t}\big(t-s_1-...-s_{k-2}\big)^2-\lambda^k e^{-\lambda t}\frac{(t-s_1-...-s_{k-2})^2}{2}=\lambda^k e^{-\lambda t}\frac{(t-s_1-...-s_{k-2})^2}{2}\,.
$$
And so forth. This shows that (1) is
$$
\mathbb P(N_t=k)=\lambda^ke^{-\lambda t}\frac{t^k}{k!}\,
$$
which is the well-known Poisson distribution with parameter $\lambda\, t$.
Conversely, when we start with a Poisson process $N_t$ then its
first jump time has an exponential distribution function:
\begin{align}
\mathbb P(\tau_1\le t)=\mathbb P(N_t\ge 1)=\sum_{k=1}^\infty\frac{(\lambda t)^k}{k!}\,e^{-\lambda t}
=\Big(\underbrace{\sum_{k=0}^\infty\frac{(\lambda t)^k}{k!}}_{\displaystyle e^{\lambda t}}-1\Big)\,e^{-\lambda t}=1-e^{-\lambda t}\,.
\end{align}
To find the distribution of $\sigma_2=\tau_2-\tau_1$ we look at the joint distribution of $\tau_1$ and $\tau_2$ first: for $t_1\le t_2\,,$
\begin{align}
&\mathbb P(t_1<\tau_1,\tau_2\le t_2)=\mathbb P(N_{t_1}<1,N_{t_2}\ge 2)=\mathbb P(N_{t_1}=0)-\mathbb P(N_{t_1}=0,N_{t_2}\le 1)\\\tag{2}
&\stackrel{(*)}{=}e^{-\lambda t_1}-e^{-\lambda t_1}\big(1+\lambda(t_2-t_1)\big)\,e^{-\lambda (t_2-t_1)}=e^{-\lambda t_1}-e^{-\lambda t_2}-\lambda(t_2-t_1)\,e^{-\lambda t_2}\,.
\end{align}
In (*) we have used the independence of $N_{t_2}-N_{t_1}$ and $N_{t_1}\,.$
Therefore,
\begin{align}
&\mathbb P(\tau_1\le t_1,\tau_2\le t_2)=\mathbb P(\tau_2\le t_2)-
\mathbb P(t_1<\tau_1,\tau_2\le t_2)\\
&\stackrel{(2)}{=} \mathbb P(N_{t_2}\ge 2)
-e^{-\lambda t_1}+e^{-\lambda t_2}+\lambda(t_2-t_1)\,e^{-\lambda t_2}\\
&=\sum_{k=2}^\infty\frac{(\lambda t_2)^k}{k!}\,e^{-\lambda t_2}-e^{-\lambda t_1}+e^{-\lambda t_2}+\lambda(t_2-t_1)\,e^{-\lambda t_2}\\&=1-e^{-\lambda t_2}-\lambda t_2\,e^{-\lambda t_2}-e^{-\lambda t_1}+e^{-\lambda t_2}+\lambda(t_2-t_1)\,e^{-\lambda t_2}\\
&=1-\lambda t_2\,e^{-\lambda t_2}-e^{-\lambda t_1}+\lambda(t_2-t_1)\,e^{-\lambda t_2}\,.
\end{align}
Consequently,
$$\tag{3}
\mathbb P\Big(\tau_1\in[t_1,t_1+dt],\tau_2\in[t_2,t_2+dt]\Big)=\lambda^2e^{-\lambda t_2}1_{\{t_1\le t_2\}}\,dt_1\,dt_2\,.
$$
It follows that $\sigma_2$ has an exponential distribution:
\begin{align}
&\mathbb P(\sigma_2\le t)=\mathbb P(\tau_2\le t+\tau_1)=\int_0^\infty\int_{t_1}^{t+t_1}\lambda^2e^{-\lambda t_2} \,dt_2\,dt_1=\lambda \int_0^\infty e^{-\lambda t_1}-e^{-\lambda(t+t_1)}\,dt_1\\
&=1-e^{-\lambda t}\,.
\end{align}
The remaining task is now to show this for all $\sigma_n$. I suspect that the joint distribution of $\tau_{n-1}$ and $\tau_n$ is also given by (3).
