How do I apply the Pumping Lemma to prove that this language is not regular?
Assume S is regular.
The pumping lemma says: There is a pumping length n. And if you take any string w in S with length &gt; n, then w contains a non-empty substring in the first n symbols that can be repeated 0, 2, 3, 4, ... times, and the result is in the language. (Why is that? A regular language has a finite state machine. If that state machine has n states, then while you are processing the first n symbols, you must enter one state twice, say the state X. So you could take the symbols that took you from X back to X, and remove them, or duplicate them, or repeat them a million times, and your string would still be in the language). So w = xyz, xy has length ≤ n, y has length ≥ 1, and xz, xyz, xyyyyyyyyyz etc. are all in the language.
Take any string v with length k &gt; n which doesn't end in a. The string w = v a^(7k) is in the language. When we split w into xyz, since the length of xy must be at most n, but v has a length k &gt; n, we actually split into $w = xyz a^{7k}$.
The pumping lemma says that $xy^mz a^{7k}$ is in the language for every y. But for m ≠ 1, 7k is not seven times the length of $xy^mz$ so it is not in the language, therefore S is not regular.