# 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?

• I figured out that the powers do not multiply rather concatinate. So my previous question is invalid. So my new problem is how do I prove it then? – thokthak Mar 20 '12 at 5:44
• Please check out our reference question and edit your question accordingly; where are you stuck, specifically? – Raphael Aug 14 '12 at 23:39