Why is the probability of return after exactly $2n$ steps in a simple symmetric random walk on $\mathbb{Z}$ given by
$\mathbb{P}(X_{2n} = i|X_{2n-1} \neq i,..., X_{1}\neq i,X_0 = i) = \frac{2n \choose n}{2^{2n}(2n-1)}$, $(*)$
as in this answer?
The probability of return after $2n$ steps is very clearly
$\mathbb{P}(X_{2n} =i|X_0=i) = $ $2n\choose n$ $p^nq^n = $ $2n\choose n$ $\frac{1}{2^{2n}}$ for a symmetric random walk with $p = q=1/2$.
But where does the $1/(2n-1)$ factor in $(*)$ come from?
Without loss of generality, let $i=0$. Note that you are dealing with a first passage time (i.e., the first time your random walk hits $0$), and therefore, it is not sufficient to just consider the current state at time $2n$.
In the second equation (the one not marked), you are calculating the probability of the following: in the first $2n$ steps, what is the probability that we returned to the origin (point $0$) at time $2n$, regardless of whether we have returned previously? To illustrate this point, here's a counter example of why your proposed answer is incorrect. Suppose we start at $0$, and the random walk makes the following steps: $RLRRLL$, then the first return time is 2 rather than $2(3) = 6$. Your proposed answer would be calculating probabilities of this form.
The $(2n-1)$ is to correct for this. Since $2n \choose n$ does not correct for the above (where it is possible to have passed by the origin in $2k$ steps for some $k < n$). Knowing this form, and by the above construction, one can verify by induction, but this feels like cheating. Lets try to derive this correction constant from scratch (using some heuristic arguments at times to avoid unnecessary complications).
For convenience, let us introduce some notation. Let random variable $\tau_0$ be the first return time to $0$ (that is, $X_{\tau_0} = 0$). The probability $\mathbb{P}(X_{2n} =i|X_0=i)$ can be rewritten as, $$ \mathbb{P}(X_{2n} =i|X_0=i) = \mathbb{P}(\tau_0 = 2n \text{ and } \tau_0 \leq 2n), $$ and similarly, $$ \mathbb{P}(X_{2(n-1)} =i|X_0=i) = \mathbb{P}(\tau_0 = 2(n-1) \text{ and } \tau_0 \leq 2(n-1)). $$ Now, since the random walk follows the Markov property (the future only depends on the immediate past), we claim that the random walk 'restarts' every time it hits $0$. That is, every time the random walk hits $0$, we view this random walk as a new random walk with origin $0$. Therefore, using this line of reasoning, $$ \mathbb{P}(\tau_0 = 2n) = \mathbb{P}(\tau_0 = 2(n-1) \text{ and } \tau_0 \leq 2(n-1)) - \mathbb{P}(\tau_0 = 2n \text{ and } \tau_0 \leq 2n). $$ Rewriting $$ {2(n-1) \choose n-1} \frac{1}{2^{2(n-1)}} = {2n \choose n} \frac{n}{2n-1}\frac{1}{2^{2n}} 2, $$ and substituting this into the above equation yields the desired result.
