# Ordinal exponentiation of limit ordinal

Prove that If $a$ is a limit ordinal and $p$ and $q$ are finite ordinals, then $(ap)^q$ = $a^q$$p^q$.

In my book, the right side of the equality is written as $a^qp$, but i think its typo since equality does not hold for the case $q=0$ and $p\ge1$.

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The book's version is the correct one for $q>0$. –  Chris Eagle May 28 '12 at 19:13

The book is correct, except in the case that $q=0$ and $p\neq 1$. One result you might want to prove is that for any limit ordinal $\lambda$ and any non-$0$ finite ordinal $p$, we have $p\cdot\lambda=\lambda$. From that, associativity of multiplication, and the fact that $0\cdot\alpha=0=\alpha\cdot 0$ for all $\alpha$, the book's equality follows readily by induction for non-$0$ finite $q$.