Let $M_t$ be a finite, discrete martingale with filtration $\{\mathcal F_t : t=0,...,T\}$. Prove that, for $0\le s < t \le T$,
$$\Bbb E[M_t|\mathcal F_s] = M_s $$
I attempt to prove this by induction. For a fixed $s$, the martingale property yields $\Bbb E[M_{s+1}|\mathcal F_s] = M_s$, this furnishes the base case. Pick an integer $n\ge 1$ so that $\mathcal F_s \subset \mathcal F_{s+n}$ and assume that the result holds for $s+n$, then
$$
\begin{align}
\Bbb E [ M_{s+n+1}|\mathcal F_{s+n}]&= M_{s+n} \tag{1} \\ \Bbb E[ \Bbb E [M_{s+n+1}|\mathcal F_{s+n}]|\mathcal F_{s}]= \Bbb E[M_{s+n+1}|\mathcal F_s] &= \Bbb E [M_{s+n}|\mathcal F_s] \tag{2} \\ &= M_s \tag{3}
\end{align}
$$
Where $(1)$ is simply the martingale property, $(2)$ follows from $(1)$ and the tower property of conditional expectation and $(3)$ follows from $(2)$ and the induction hypothesis. Since this induction argument can be repeated for every $s$, the result follows.
Unfortunately, a hint was given to use induction for a fixed $s$ to prove that the result held for all $t>s$, and then use induction again to show that this holds for all $s$. It was also hinted at to use the fact that martingales have constant mean.
The problem is I do not see where all of these other arguments fit in. This leads me to believe that my argument is wrong, but as is often the case when one is wrong, I cannot see how. The base case and $(1)$ are given. The tower property for $(2)$ holds because we are told that the $\mathcal F$'s form a filtration so that $\mathcal F_s$ is coarser than $\mathcal F_{s+n}$. Where is the mistake?
First let $s=0$, then by definition $\mathbb E[M_1\mid \mathcal F_0] = M_0$. Suppose $\mathbb E[M_t\mid\mathcal F_0] = M_0$ for some $1\leqslant t\leqslant T-1$. Then $$ \mathbb E[M_{t+1}\mid \mathcal F_0] = \mathbb E[\mathbb E[M_{t+1}\mid\mathcal F_t]\mid \mathcal F_0] = \mathbb E[M_t\mid\mathcal F_0] = M_0. $$ Suppose now that $\mathbb E[M_t\mid \mathcal F_s]=M_s$ for some $0\leqslant s\leqslant T-2$ and $s<t\leqslant T-1$. Then $$ \mathbb E[M_{t+1}\mid\mathcal F_s] = \mathbb E[\mathbb E[M_{t+1}\mid\mathcal F_t]\mid \mathcal F_s] = \mathbb E[M_t\mid\mathcal F_s]=M_s. $$ It follows that $\mathbb E[M_t\mid\mathcal F_s]=M_s$ for $0\leqslant s<t\leqslant T$.