A weak Goodstein sequence starting at $m>0$ is a sequence $m_0,m_1,\cdots$ of natural numbers defined by $m_0=m$. To obtain $m_{k+1}$ form $m_k$ (as long as $m_k\neq 0$), write $m_k$ in base $k+2$, increase the base by $1$ (to $k+3$), and subtract $1$. For example, the weak Goodstein sequence starting at $m=21$ is as follows:
$m_0=21=2^4+2^2+2^0$
$m_1=3^4+3^2+3^0-1=90$
$m_2=4^4+4^2-1=271$
$m_3=5^4+5^1\cdot 3+5^0\cdot 3-1=642$
Theorem: For each $m>0$, the weak Goodstein sequence starting at $m$ eventually terminates with $m_n=0$ for some $n$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $m_0,m_1,\cdots$ be the weak Goodstein sequence starting at $m$. Its $a^{\rm th}$ term is written in base $a+2$: $$m_a=(a+2)^{b_1}k_1+\cdots+(a+2)^{b_n}k_n$$
Consider the ordinal $\alpha_a=\omega^{b_1}\cdot k_1+\cdots+\omega^{b_n}\cdot k_n$ obtained by replacing $a+2$ by $\omega$. We have $$\begin{align}m_{a+1}&=(a+3)^{b_1}k_1+\cdots+(a+3)^{b_n}k_n-1\\&=(a+3)^{b_1}k_1+\cdots+(a+3)^{b_{n-1}}k_{n-1}+\left((a+3)^{b_n}k_n-1\right)\\&=(a+3)^{b_1}k_1+\cdots+(a+3)^{b_{n-1}}k_{n-1}+(a+3)^{b'_n}k'_n\end{align}$$ where $(a+3)^{b_n}k_n-1=(a+3)^{b'_n}k'_n$.
It follows that $b_n>b'_n$ and that $\alpha_a>\alpha_{a+1}=\omega^{b_1}\cdot k_1+\cdots+\omega^{b_{n-1}}\cdot k_{n-1}+\omega^{b'_n}\cdot k'_n$. Then $\alpha_0>\alpha_1>\cdots>\alpha_a>\cdots$ is a decreasing sequence of ordinals.
If $(m_a \mid a\in\Bbb N)$ did not go to $0$, it would not terminate and thus be infinite. Consequently, $(\alpha_a\mid a\in \Bbb N)$ would be infinite. This contradicts the fact that $<$ is a well-founded relation on ordinals.
