Let $p_{xx}^{(n)}$ be the probability of returning to $x$ in $n$ steps. Let $f_{xx}^{(n)}$ be the probability of first returning to $x$ in $n$ steps.
The period of a state $x$ is defined to be $gcd\{n\ge1|p_{xx}^{(n)}>0\}$. An equivalent definition of the period of $x$ is $gcd\{n\ge1|f_{xx}^{(n)}>0\}$.
Intuitively, $gcd\{n\ge1|p_{xx}^{(n)}>0\}=gcd\{n\ge1|f_{xx}^{(n)}>0\}$ but how to prove it rigorously?
Clearly, if $f_{xx}^{(n)} > 0$, then $p_{xx}^{(n)} > 0$, because if you return to $x$ after $n$ steps for the first time, in particular you return to $x$ after $n$ steps. So if $d$ divides all $n$ such that $p_{xx}^{(n)} > 0$, then $d$ divides all $n$ such that $f_{xx}^{(n)} > 0$.
Going in the other direction, we know that if the Markov chain returns to $x$ after $n$ steps, we can write $n = n_1 + n_2 + \dots + n_k$ where each $n_i$ is an interval between visits to $x$, and so $f_{xx}^{(n_i)} > 0$ for all $i$. If $d$ divides all $n$ such that $f_{xx}^{(n)} > 0$, then $d$ divides each of $n_1, n_2, \dots, n_k$. So $d$ divides $n_1 + n_2 + \dots + n_k$, which means that $d$ divides all $n$ such that $p_{xx}^{(n)} > 0$.