If $\{Z_n, n \geq 1\}$ is a martingale show that, for $1 \leq k < n$. $E[Z_n|Z_1,...,Z_k] = Z_k$

Question

If $\{Z_n, n \geq 1\}$ is a martingale show that, for $1 \leq k < n$. \begin{equation} E[Z_n|Z_1,...,Z_k] = Z_k \end{equation}

I know that $E[Z_{k+1}|Z_1,...,Z_k] = Z_k$ and $Z_{k+1} = E[Z_{k+2}|Z_1,...,Z_{k+1}]$. Plugging the second into the first gives $E[E[Z_{k+2}|Z_1,...,Z_{k+1}]|Z_1,...,Z_k] = Z_k$. If i can somehow show that this implies $E[Z_{k+2}|Z_1,...,Z_k] = Z_k$ then the rest should follow from induction. But I'm not sure why this would be the case.

Answer 1

$$E[Z_{k+2}|Z_1,...,Z_k]$$ $$=E[E[Z_{k+2}|Z_1,...,Z_{k+1}]|Z_1,...,Z_k]$$$$=E[Z_{k+1}|Z_1,...,Z_k]=Z_k.$$ The first step is the tower property of conditional expectation.