Let $(X_n)_{n\in\mathbb{N}_0}$ be a Markov chain with state space $E$. For $i,j\in E$ set $$ h_i(j):=\mathbb{P}_i(H(j)<\infty):=\mathbb{P}(H(j)<\infty|X_o=i), $$ where $H(j)\colon \Omega\to\mathbb{N}_0\cup\left\{\infty\right\}, \omega\mapsto\inf\left\{n\geq 0: X_n(\omega)=j\right\}$.
Show that for $i,j,k\in E$ it is $$ h_i(k)\geq h_i(j)\cdot h_j(k). $$
Ok, here is what I made up to now using the definition of conditional probability and the strong Markov property:
\begin{align} h_i(k)&=\mathbb{P}_i(H(k)<\infty)=\sum_{n=0}^{\infty}\mathbb{P}_i(H(k)=n)\\ &\geq\sum_{n=0}^{\infty}\mathbb{P}_i(H(k)=n,H(j)\leq n)\\ &=\sum_{n=0}^{\infty}\sum_{\ell=0}^{n}\mathbb{P}_i(H(k)=n,H(j)=\ell)\\ &=\sum_{n=0}^{\infty}\sum_{\ell=0}^{n}\mathbb{P}_i(H(k)=n|H(j)=\ell)\cdot\mathbb{P}_i(H(j)=\ell)\\ &=\sum_{n=0}^{\infty}\sum_{\ell=0}^n \mathbb{P}_j(H(k)=n-\ell)\cdot\mathbb{P}_i(H(j)=\ell) \end{align}
Now because of absolute convergence one can exchange the sums. But I am not sure how to adapt the index of the sums; I think it is \begin{align} \sum_{n=0}^{\infty}\sum_{\ell=0}^n \mathbb{P}_j(H(k)=n-\ell)\cdot\mathbb{P}_i(H(j)=\ell)&=\sum_{\ell=0}^{\infty}\sum_{n=\ell}^{\infty}\mathbb{P}_j(H(k)=n-\ell)\cdot\mathbb{P}_i(H(j)=\ell)~~~~(*)\\ &=\sum_{\ell=0}^{\infty}\mathbb{P}_i(H(j)=\ell)\sum_{n=\ell}^{\infty}\mathbb{P}_j(H(k)=n-\ell)\\ &=\sum_{\ell=0}^{\infty}\mathbb{P}_i(H(j)=\ell)\sum_{n=0}^{\infty}\mathbb{P}_j(H(k)=n)\\ &=\mathbb{P}_i(H(j)<\infty)\cdot\mathbb{P}_j(H(k)<\infty)\\&=h_i(j)\cdot h_j(k). \end{align}
1.) Am I right resp. is my proof okay?
2.) I did the index adaption in line $(*)$ more intuitively. Could you please tell me how I formally get from $$ \sum_{n=0}^{\infty}\sum_{\ell=0}^n\ldots\text{ to }\sum_{\ell=0}^{\infty}\sum_{n=\ell}^{\infty}\ldots? $$
With greetings