The Problem:
Consider a Markov chain with the $N+1$ states $0,1,...,N$ and transition probabilities \begin{align*} P_{ij} &= {N \choose j}\pi_{i}^{j}(1 - \pi_i)^{N-j}, \hspace{5mm} 0 \le i, j \le N, \\ \pi_i &= \frac{1 - e^{-2ai/N}}{1 - e^{-2a}}, a > 0. \end{align*} Note that $0$ and $N$ are absorbing states. Verify that $\exp(-2aX_t)$ is a martingale $[$or, what is equivalent, prove the identity $\mathbb{E}[\exp(-2aX_{t + 1})|X_t] = \exp(-2aX_t)$$]$, where $X_t$ is the state at time $t$ $(t = 0,1,2,...)$. Using this property show that the probability $P_N(k)$ of absorption into state $N$ starting at state $k$ is given by $$ P_N(k) = \frac{1 - e^{-2ak}}{1 - e^{-2aN}}. $$
Hint: Use the fact that absorption into one of the states $0$ or $N$ in finite time occurs with certainty and that the relations $$ \mathbb{E}[\exp(-2aX_0)] = \mathbb{E}[\exp(-2aX_n)] = P_N(k) \exp(-2aN) + (1 - P_N(k)) $$ hold (justify this).
My Progress:
As the hint suggests, I started by trying to prove the identity $$ \mathbb{E}[\exp(-2aX_0)] = \mathbb{E}[\exp(-2aX_n)]. $$ Well, we can write the LHS as $$ \mathbb{E}[\exp(-2aX_0)] = \sum_{j=0}^{N}\exp(-2aj)\mathbb{P}(X_0 = j). $$ But this doesn't seem right because it seems a bit strange to be working with the pmf of the random variable $X_0$. Nevertheless, if I know that if I can get the same expression for $\mathbb{E}[\exp(-2aX_n)]$, I should be in good shape. However, this seems to a bit more unwieldy if I use the same approach. Indeed, $$\mathbb{E}[\exp(-2aX_n)] = \sum_{j=0}^{N}\exp(-2aj)\mathbb{P}(X_n = j),$$ where (I'm assuming), for each $j$, $$ \mathbb{P}(X_n = j) = \sum_{i=0}^{N}P_{ij}^{n}\mathbb{P}(X_0 = i). $$ That is, I'm just using the marginals and the Markov property to find the pmf of $X_n$. Again, this doesn't seem right to me; but I can't think of anything else to do...
What am I missing here?