# Limit ordinal in the exponent [duplicate]

How would you prove that $$n^{\gamma} = m^{\gamma}$$ for every limit ordinal $\gamma$ and $n,m$ finite ordinals?

It's a rather short solution problem, but I can't construct any slick answer for it. I know very little about ordinal exponentiation, just that $\alpha^{\beta +1} = \alpha^{\beta}\cdot\alpha$ and that if there's a limit ordinal in the exponent we take the $\sup$.

## marked as duplicate by Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 22 '16 at 19:35

You need to assume that $m,n>1$.
First show that it’s true for $\gamma=\omega$. Then show that if it’s true for some limit $\gamma$, it’s true for $\gamma+\omega$; this is actually a bit like the first step. Finally, show that if it’s true for every limit ordinal less than $\gamma$, and $\gamma$ is a limit of limit ordinals, then it’s true for $\gamma$ as well.