Let $(X_0, X_1, ...)$ be an irreducible, recurrent Markov chain and $T_k$ be the first-passage time to $k\in S$, i.e. $T_k = \inf\{n\geq1 : X_n = k\}$.
My aim is to show that, for $i \in S$, $$\Sigma_{r=0}^{\infty}\mathbb{P}_k(X_r=i, T_k \geq r+1)=\Sigma_{m=1}^{\infty}\mathbb{P}_k(X_m=i,T_k\geq m).$$ For context, see Probability (Grimmett), Page 233. I began by separately considering the cases where $i=k$ and $i \neq k$. For the latter case, we see the first summands are zero and the rest are equivalent. For the former, we have the first summand as $$\mathbb{P}_k(X_0=k, T_k \geq 1)= \mathbb{P}_k(X_1=k,T_k\geq 1)=1.$$ For the other summands, the terms from the first summation are zero, since $X_r = k$ implies that $T_k \leq r$ when $r>0$. Hence it remains to show that $$\mathbb{P}_k(X_m=k,T_k\geq m)=0$$ for all $m >1$. I'm having a hard time doing this. Since $X_m = k$ implies that $T_k \leq m$, we have $$\mathbb{P}_k(X_m=k,T_k\geq m)= \mathbb{P}_k(X_m=k,T_k= m).$$ Why is this zero?
The case $i\neq k$ is clear as the OP said.
For the case $i=k$, $\mathbb P_k(X_r=k, T_k\geq r+1)=0$ for $r\geq 1$ but for $r=0$ we have $\mathbb P_k(X_0=k, T_k\geq 1)=1.$ Finally, we have \begin{aligned} RHS&=\sum_{m=1}^\infty\mathbb P_k(X_m=k, T_k\geq m)\\&=\sum_{m=1}^\infty P_k(X_m=k, T_k=m)\\&=\sum_{m=1}^\infty P_k( T_k=m)\\&=1=LHS. \end{aligned} The proof is complete now.