Consider a Markov chain $\left(X_{n}\right)_{n \geq 0}$ on the state space $S=$ $\{1, \ldots, m+1\}, m \geq 1$, with transition probability matrix of dimension $(m+1) \times(m+1)$
$$ P=\left(\begin{array}{cccc} 0 & \frac{1}{m} & \ldots & \frac{1}{m} \\ 1 & 0 & \ldots & 0 \\ \vdots & \vdots & & \vdots \\ 1 & 0 & \ldots & 0 \end{array}\right) $$
Let $T_{i}(1)=\min \left\{n>0 \mid X_{n}=i\right\}$ be the first time in state $i$ after time zero. Show that
$p_{11}^{(2 n)}=1$ and $p_{11}^{(2 n+1)}=0$ for $n \in \mathbb{N}$.
I am really new to working with transition matrices. From my understanding the notation $p_{11}^{2n}$ is the probability of going from state $1$ to state $1$ in $2n$ steps which would be the first entry, i.e staying in the same first state. However, I have no idea on how I can calculate this for odd and even powers using the TPM, $P$.
Computing $$P^2 = \left(\begin{array}{cccc} 1 & 0 & \ldots & 0 \\ 0 & \frac1m & \ldots & \frac1m \\ \vdots & \vdots & & \vdots \\ 0 & \frac1m & \ldots & \frac1m \end{array}\right)\quad \text{and}\quad P^3 = \left(\begin{array}{cccc} 0 & \frac1m & \ldots & \frac1m \\ 1 & 0 & \ldots & 0 \\ \vdots & \vdots & & \vdots \\ 1 & 0 & \ldots & 0 \end{array}\right) = P$$
How do I by induction show that $P^{2n} = P^2$ and $P^{2n+1} = P$ which proves my claim?
Start by showing $P^4 = P^2$. Then for $n \geq 2$:
$$P^{2n} = P^4 P^{2(n-2)} = P^2P^{2(n-2)} = P^{2(n-1)}$$
and so you can apply the induction hypothesis. For the odd terms do the same.