Let $q(i,i+1)$ is a jump rate of a Markov chain from a state i to a state i+1.
Let $q(i,i+1)=\lambda i^{p}$ which has the form $q(i,i+1)=\lambda u(i,i+1)$ where $u(i,i+1)$ is the transition probabilities of a Markov chain at state i to an adjacent state i+1.
Clearly, for a Markov chain at state i, there is a finite time interval $T_{n}=t_{1}+\cdot \cdot \cdot+t_{n}$ that has elapsed.
It is claimed in the text Essential of Stochastic Process that the inter-arrival times $t_{i}$ is exponential with rate $n^{p}$
I am able to understand that the inter-arrival times between events are exponential. This follows from a definition of poisson process.
But why does it has a rate $n^{p}$?
Edit: Excerpt added.
When you're at state $n$, the only possibility is to go to state $n+1$, and the transition rate from $n$ to $n+1$ is $\lambda n^p$, so the distribution of the waiting time until we go from $n$ to $n+1$ is $t_n \sim \operatorname{Exp}(\lambda n^p)$ by definition.
In the text you're quoting, the $\lambda$ disappears mysteriously here, but reappears later on in the sum, which I assume is just a typo. We should have $E[t_n] = \frac1{\lambda n^p}$ and $E[T_n] = \frac1\lambda \sum_{m=1}^n \frac1{m^p}$.