Do self-loops in DFA cause infinite languages?

A true/false question: If a DFA $M$ contains a self-loop on some state $q$, then $M$ must accept an infinite language.

The answer is "false". I've read this question, but I'm still wondering why $M$ does not necessarily accept an infinite language. Isn't the language $b^*$ infinite? Don't all self-loops look like $b^*$?

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If your definition of DFA demands the graph to be complete (i.e. the transition function to be total), every minimal DFA will contain self-loops. – Raphael Jan 17 '13 at 13:39
@Raphael If a self-loop is a transition from a state to itself, then consider the minimal DFA accepting $\mathcal{L} = \{ a^{2n} \}$. This DFA contains two states and no self-loops. – Pål GD Jan 17 '13 at 15:28
@PålGD Oh, right; my argument only seems to work for languages that don't contain a word for every possible prefix. – Raphael Jan 17 '13 at 16:52

To generalize, any state that either isn't reachable from the start state, or that cannot reach any accept state can be have self-loops without any immediate consequence. A very easy exercise: Consider the alphabet $\Sigma = \{a\}$ and construct a DFA that accepts the language $\mathcal{L} = \{a\}$. – Pål GD Jan 17 '13 at 10:53