Question on proof of Goodstein's Theorem I recently read the ordinal-based proof of Goodstein's Theorem, saying that all Goodstein sequences do terminate. However, I did not see why the crucial part worked.
To my knowledge, the proof involves the construction of a parallel ordinal sequence to each Goodstein sequence $G(m)(n)$, $P(m)(n)$, so that $\exists G(m)(n)\Longleftrightarrow\exists P(m)(n)$. Each $P(m)(n+1)$ is constructed by replacing each occurrence of $n+1$ in $G(m)(n)$ with $\omega$, so $2^2+2\rightarrow \omega^\omega+\omega$, for example.
The proof then says that since the base-changing aspect of the process going from $G(m)(n)$ doesn't affect the omegas in the corresponding $P(m)(n)$ and then we subtract one from both, the $P(m)(n)$ must become strictly smaller each step. But why is this the case?
If I have some $G(m)(n)$ which is equivalent to $2^2$, with a $P(m)(n)$ thus equal to $\omega^2$, although it is true that $3^3-1<3^3$, is it true that $\omega^2-1<\omega^2$? Is the expression "$\omega^2-1$" even well defined?
Without this strictly decreasing aspect of the $P(m)(n)$ sequence, it is not clear to me how the proof works...

As Lord Shark the Unknown points out, one first reduces the natural number by one and then changes the base, so the example above is resolved by saying $2^2\rightarrow 3^1+1$ and so $\omega^2 \rightarrow \omega+1$. Thus, since the base doesn't affect the ordinal but the subtraction of one leads to the reduction of a non-base integer, then the resulting $P(m)(n+1)<P(m)(n)$ and then the rest of the proof follows...
