Uniqueness of Cantor normal form using powers of 2 in transfinite Nim I am reading Joel Hamkins' blog post on transfinite Nim and it contains the following theorem:

Theorem. Every ordinal $\beta$ has a unique representation as a decreasing finite sum of ordinal powers of two.
$$\beta = 2^{\beta_n} + \ldots + 2^{\beta_0}, \quad \beta_n > \cdots > \beta_0$$
The proof is easy! We simply prove it by transfinite induction on $\beta$. If the theorem holds below an ordinal $\beta$, first let $2^{\alpha}$ be the largest power of two that is at most $\beta$, so that $\beta = 2^{\alpha} + \gamma$ for some ordinal $\gamma$. It follows that $\gamma < 2^{\alpha}$, for otherwise, we could have made $2^{\alpha + 1} \leq \beta$. Thus, by induction, $\gamma$ has a representation with powers of two, and so we may simply add $2^{\alpha}$ at the front to represent $\beta$. To see that the representations are unique, first establish that any power of two is equal to or more than the supremum of the finite decreasing sums of any strictly smaller powers of two. From this, it follows that any representation of $\beta$ as above must have used $2^{\alpha}$ just as we did for the first term because otherwise, it couldn’t be large enough, and then the representation of the remaining part $\gamma$ is unique by induction, and so we get uniqueness for the representation of $\beta$. QED

Can someone explain the statement in bold to me? I'm not sure what "supremum of the finite decreasing sums of any strictly smaller powers of two" means. Also, I'm not really sure how it follows from here that the representation is unique?
 A: The statement in bold
Consider something like $2^{\omega\cdot2}+2^{\omega+7}+2^{\omega}+2^{5}+2^{2}+2^{0}$. That is a sum of finitely many terms, each of which is a power of two, so it's a "finite ...sum of...powers of two". Also, it's not like $2^{0}+2^{0}+2^{0}+2^{1}$, the terms in in the sum are written in strictly decreasing order (so there are no repeats). That makes it a "finite decreasing sum of...powers of two" (FDSoPoT). Also, if we are looking at a power of $2$ that's strictly bigger than all of the terms, like $2^{\omega\cdot2+1}$, then it's a "finite decreasing sum of strictly smaller (than $2^{\omega\cdot2+1}$) powers of two".
Now, we can consider the supremum of all such things. I.e. $\sup\left\{ \delta\mid\delta\text{ is a FDSoPoT that are strictly smaller than }2^{\omega\cdot2+1}\right\}$ . The claim is that this supremum will be no greater than the original $2^{\omega\cdot2+1}$. And that that inequality holds no matter what ordinal power of $2$ we choose in place of $2^{\omega\cdot2+1}$.

Why unique?
I don't know if this was the intended argument, but I think it works:
Well, suppose $\beta$ is the least ordinal with two different representations: $2^{b_{m}}+\cdots+2^{b_{0}}$ and $2^{\beta_{n}}+\cdots+2^{\beta_{0}}$. Suppose WLOG that $b_{m}\ge\beta_{n}$. If $b_{m}=\beta_{n}$, then removing the $2^{b_{m}}=2^{\beta_{m}}$ term from each sum would give a smaller ordinal with two different representations (this is a rephrasing of the "unique by induction" part). So we must have $b_{m}>\beta_{n}$. If $\beta_{n}$ is finite, appeal to the base-$2$ representation of the finite $\beta=2^{\beta_{n}}+\cdots+2^{\beta_{0}}$. Otherwise, assume $\beta_{n}$ is infinite, choose $\delta$ so that $\beta_{n}>\delta$ but $\delta$ is not any of the $\beta_{i}$. Set $\beta'$ to be the sum $2^{\beta_{n}}+\cdots+2^{\beta_{0}}$ with an extra term of $2^\delta$ in the right spot so that it's a decreasing sum. We then have the following contradiction: $\beta=2^{b_{m}}+\cdots+2^{b_{0}}\ge2^{b_{m}}\boxed{\ge}\beta'>\beta$ where the boxed inequality follows from the previous result about suprema.
