Suppose that parts arrive at a machine according to a Poisson process with mean arrival time $\lambda t$. After finishing one part, machine goes idle for $T$ seconds, until the next part arrives.
I am asked to find the distribution of $T$ with a hint: use memoryless property of exponential distribution $\Pr(X > t + t^\prime| X > t) = \Pr(X > t^\prime)$.
I am aware that if arrival time are Poisson with rate $\lambda$, then the inter-arrival time is exponentially distributed with the same rate $\lambda$.
So $T \sim \text{exp}(\lambda)$.
How do I make use of the hint?
The thing is that you set the arrival time of the $\left(n-1\right)$th part as $s_{n-1}=0$ so in a way the memoryless property is already implied by that assumption. However, this is in a way 'inaccurate' (but actually not 'wrong' either) since we neither actually know the time taken for the $\left(n-1\right)$th arrival, $s_{n-1}$, nor that of the $n$th arrival, $s_n$.
Let $K\left(t\right)$ denotes the number of parts arriving in time $t$, $K\sim\operatorname{Po}\left(\lambda t\right)$.
Let $X_n$ denotes the time taken for the $n$th part to arrival, with respect to the very start of the process.
Let $T_n$ denotes the interarrival time between the $\left(n-1\right)$th part and the $n$th part, that is, $T_n\colon=\left(X_n-X_{n-1}\right)\mid X_{n-1}>{S_{n-1}}$ where $S_n\colon=\sum_{r=1}^{n}{s_{r}}$
So we perform the usual trick \begin{align} P\left(T_n>s_n\right)&=P\left(\left(X_n-X_{n-1}\right)>s_n\mid X_{n-1}>{S_{n-1}}\right)\\ &=P\left(X_n>S_n\mid X_{n-1}>{S_{n-1}}\right) \end{align} Applying the memoryless property \begin{align} P\left(T_n>s_n\right)&=P\left(X_n>s_n\right)\\ &=P\left(K\left(s_n\right)=0\right)\\ &=e^{-\lambda s_n} \end{align} Since this is true for all $n\in\mathbb{Z}^+$, I will drop the subscript and redefine $t\colon=s_n$. Thus, \begin{align} P\left(T>t\right) &=e^{-\lambda t} \end{align} Clearly, $T\sim\operatorname{Exp}\left(\lambda\right)$ as desired.