Yes. Here's a sketch of a direct proof.
If a problem is in , there is a nondeterministic Turing machine that decides it, and there's a polynomial such that none of 's computation paths on inputs of length take more than steps. That means that a single path can't use more than tape cells, so we can simulate a single path deterministically in polynomial space.
But we need to simulate all the paths. Well, there is a constant that depends only on the transition function of (and not on its input) such that has at most nondeterministic choices at any step. That means that there are at most different computation paths for any input of length . We can simulate all of these paths as follows. First, write out a -digit number in base- (this takes space but that's polynomial, so it's OK). Then, simulate the operation of and, at the th step of the computation, use the th digit of the number to decide which nondeterministic choice to make. If, for example, the th digit is and there are only four choices that can be made, abandon that simulation and go on to the next one.
So, now, to do the whole simulation, we start by writing out the number , simulate that path of , increment the number, simulate the next path, and so on, until we reach the number where every digit is . We've now simulated every possible computation path, and we've done it in time about , using space about . That's exponential time and polynomial space, as required.