Suppose I have an irreducible finite state Markov chain $X_0,X_1,\cdots$. For an state $x$ and any $n$, return time is defined as $\tau_x^n=\inf\{ t> \tau_x^{n-1}\mid X_t = x\}$, and $\tau_x^0 = 0$. Why is it true that the differences of return time to $x$ are iid? For example, why it is true that
$$\tau_x^4 - \tau_x^3 \text{ and }\tau_x^3 - \tau_x^2$$
are iid?
I am given that $h\left(X_0,\cdots,X_{\tau_x^1} \right)$ and $h\left(X_{\tau_x^1},\cdots,X_{\tau_x^2}\right)$ are iid for any $h$, where $\tau$ is stopping time (without proof though). Since return time is stopping time, maybe I can use this result? But it's not clear how to expess $\tau_x^{n} - \tau_x^{n-1}$ into this form. I know that by definition, $\tau_x^3$ depends on
$$X_{\tau_x^2 + 1},X_{\tau_x^2 + 2},\cdots$$
and that $\tau_x^2$ depends on $X_{\tau_x^1 + 1},\cdots$. But I don't know how to continue.
My attempt: For any $n$, we have
\begin{align*} \mathbb{P}\left(\tau_x^{n+1} - \tau_x^{n} = t\right) &= \mathbb{P}\left(\tau_x^{n+1} = \tau_x^n + t \right)\\ &= \mathbb{P}\left(\inf\{s> \tau_x^n \mid X_s = x \} = \tau_x^n + t \right)\\ &= \mathbb{P}\left(X_{\tau_x^n +1} \ne x, X_{\tau_x^n + 2} \ne x,\cdots,X_{\tau_x^n + t} = x \right) \\ &= \mathbb{P}\left(Y_1 \ne x, Y_2\ne x,\cdots, Y_t = x \right) &&\text{define }Y_t := X_{\tau_x^n + t} \end{align*}
On the other hand,
$$\mathbb{P}\left(\tau_x^1 = t\right) = \mathbb{P}\left(X_1\ne x, X_2 \ne x,\cdots,X_t = x\right)$$
By the Markov property, it "starts afresh" each time it returns to $x$. In other words, the distribution of $\tau_x^n - \tau_x^{n-1}$ is the same as the distribution of $\tau_x^1$ because it measures the exact same thing: the length of a random sojourn that leaves $x$ and then returns to $x$.
The independence is clear because whatever happens on the first sojourn does not affect what happens on the next, etc.