Pumping Lemma: is it valid to "multiply the product of powers" in this case?

I need to show that

$\qquad \displaystyle S = \{(10^p)^m \mid p \geq 0, m \geq 0\}$

is not a regular language using pumping lemma.

Can I multiply the product of the powers and express it to: $S = \{ 1^m 0^{pm} \mid \dots \}$ and apply the pumping lemma where I pump 1's then say that the language doesn't accept the new string?

You're right, the superscript in this context means concatenation.

To prove a language isn't regular, you use the fact that any regular language can be "pumped". First, find out what "pumping" a word means (I can't do your homework for you), and then show that your language can not be pumped, and thus, can not be regular. Basically you will take a sufficiently long word that belongs to your language, and show that it cannot be broken up and pumped in such a way that it satisfies the pumping lemma.

Note that the converse of the pumping lemma is not always true: that is, if a language does NOT satisfy the pumping lemma, it may still be non-regular. For this reason, the pumping lemma is used to exclude a language from a set of languages, and not to include a language in a set of languages (such as regular). In other words, satisfying the pumping lemma is a necessary, but not sufficient, condition for a language to be regular.

StackExchange Computer Science Q#3190, answer score: 7