Answer
M accepts $0{n+1} 1^{n+1}$ it also accepts $0^{n+k+1} 1^{n+1}$, which
is a contradiction.
Work Step by Step
Suppose that M is a finite-state automaton that accepts the set of bit strings containing an equal number of 0s and 1s.
Suppose M has n states. Consider the string $0^{n+1}1^{n+1}$.
By the pigeonhole principle, as M processes this string, it must encounter the same state more than once as it reads the first n+1 0s;
so let s be a state it hits at least twice. Then k 0s in the input takes M from state s back to itself for some positive integer k.
But then M ends up exactly at the same place after reading $0^{n+1+k} 1^{n+1}$ as it will after reading $0^{n+1} 1^{n+1}$.
Therefore, because M accepts $0{n+1} 1^{n+1}$ it also accepts $0^{n+k+1} 1^{n+1}$, which
is a contradiction.
