Why does a finite, irreducible and aperiodic Markov chain with a doubly-stochastic matrix P have a uniform limiting distribution?

The theorem is "If a transition matrix for an irreducible Markov chain with a ﬁnite state space S is doubly stochastic, its (unique) invariant measure is uniform over S."

If a Markov Chain has a doubly-stochastic transition matrix, I read that its limiting probabilities make up the uniform distribution, but I do not quite understand why.

I have been trying to come up with, and locate, an understandable proof for this. But the proofs I find all gloss over details I don't understand, like proposition 15.5 here (why does it work to just use the [1,...1] vectors?) Could someone point me to (or write) a more simple/detailed proof?

(Though not part of anything I will hand in at school, it is part of a course I'm taking so I guess I'll tag it with homework in either case.)

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Perron-Frobenius. –  cardinal Sep 21 '12 at 10:30
@cardinal Why not make it an answer with a little elaboration? –  Michael Chernick Sep 21 '12 at 16:34
You are missing the necessary conditions that the Markov Chain is irreducible and not periodic. These can be combined into the condition that for some $n$, every entry of $P^n$ is positive. There are finitely many, so say all are at least $c$. You can bound the convergence rate in terms of $c$. –  Douglas Zare Sep 21 '12 at 17:45
You're right, Douglas. I have now copied the proposition in the linked PDF verbatim to avoid any confusion. Thanks. –  Christian Jonassen Sep 21 '12 at 23:21