Let $\{X_t:t=0,1,2,\ldots \}$ be a homogeneous Markov chain with state space $\mathcal{S}=\{1,\ldots,n\}$ and transition matrix $$p_{ij}=\binom{n}{j}\left(\frac{i}{n}\right)^j\left(\frac{n-i}{n}\right)^{n-j}\;\;\text{ for }i,j=1,\ldots,n.$$ I want to show that the process is a Martingale with respect to the natural filtration $\{\mathcal{F}_t\}$. So we compute $$ E[X_{t+1}\mid \mathcal{F}_t](\omega)=E[X_{t+1}\mid \sigma(X_t)](\omega)=E[X_{t+1}\mid X_t]\circ X_t(\omega),$$ where $$E[X_{t+1}\mid X_t](j)=E[X_{t+1}\mid X_t=j],\;\;\;\;\text{ for all } j\in\mathcal{S}.$$ (I'm expecting $E[X_{t+1}\mid X_t=j]$ to be equal to $j$ for the process to be a Matringale). So I computed $$E[X_{t+1}\mid X_t=j]=\frac{E[1_{\{X_t=j\}}X_{t+1}]}{ \mathbb{P}\{X_t=j\}} = \frac{\sum_{i=1}^ni\cdot\mathbb{P}(X_{t+1}=i\mid X_t=j)}{\sum_{i=1}^n p_{ij} \mathbb{P}(X_0=i)}=\frac{\sum_{i=1}^n i\cdot p_{ij}}{\sum_{i=1}^n p_{ij} \mathbb{P}(X_0=i)}$$
And I got stuck on the last part. First of all, is it true that $$E[1_{\{X_t=j\}}X_{t+1}]=\sum_{i=1}^ni\cdot\mathbb{P}(X_{t+1}=i\mid X_t=j) ?$$
If yes how to proceed? I know I still need to replace $p_{ij}$ with it's appropriate value but the computation seems like a mess!
I assume that $p_{0j}=1\{j=0\}$ and $p_{nj}=1\{j=n\}$. Then $\mathsf{E}[X_{t+1}\mid X_t=0]=0$, $\mathsf{E}[X_{t+1}\mid X_t=n]=n$, and for $0<i<n$, $$ \mathsf{E}[X_{t+1}\mid X_t=i]=\sum_{j=1}^nj\cdot p_{ij}=n\cdot\frac{i}{n}=i $$ because $X_{t+1}\mid X_t=i\sim \text{Bin}(n,i/n)$.