Call a space between two consecutive elements a divider. For any particular $t \in M$, occurrences of $t$ in the sequence divide the dividers into odd and even (such that there are respectively odd/even $t$s before the divider). Note that the segment between two dividers of the same parity has an even number of $t$s in it. In this way a set of $n$ parities corresponding to each $t$ is assigned to every divider. Since there are $2^n+1$ dividers, some two of them have the same parity corresponding to every $t$. Then the product of elements of the segment between them is a perfect square.