Given M = (Q,Σ,δ,q0,F) a DFA with n states. Prove:
The languge T(M) is infinite iff contains a string with lenght t, where n ≤ t < 2n.
Ok, it's intuitive for me, I can understand to get a string with length > number of states I need a loop, but how to prove it formally?
Using $\delta$-hat, could be an idea?
HINT: Consider the sequence of states visited when $M$ reads a word $w$ of length $n$ or more: it must intersect itself. That is, there must be some state that it visits twice. That means that the path contains a loop. Now you can modify the input word so that the DFA goes around that same loop any number of times while still ending in the same state as it does for $w$. Thus, if it accepts $w$, it accepts infinitely many other words. This is the basic idea; you should try to make it a bit more formal.
All that remains at that point is to show that if $|w|\ge 2n$, you can shorten $w$ to a word whose length is between $n$ and $2n-1$ inclusive that sends $M$ to the same end state. You can do this by showing that some subsequence of $w$ sends $M$ around a loop, so that this subsequence can be removed without changing the end state. As long as you have at least $2n$ symbols in the word, you can continue doing this. Finally, show that the substring sending $M$ around a loop once can have length at most $n$, so that removing it cannot drop the total length of the word below $n$.