Let $n$ be a natural number. When constructing the Goodstein sequence $(n)_{k}$, we start with $(n)_{1}=n$ written in complete base $2$, and we proceed recursively. That is, if $(n)_{k}$ has been defined and it is not $0$, $(n)_{k+1}$ is obtained from $(n)_{k}$ by bumping the base by $1$ in the complete base $k+1$ representation of $(n)_{k}$ and then subtracting $1$.
Now, let $n\ge 4$ and let $2^{m_{1}}+\cdots 2^{m_{r}}$, with $m_{1}>\cdots >m_{r}\ge 0$ be the standard base $2$ representation of $n$. As we write down the sequence $(n)_{1}$, $(n)_{2}$, etc. we notice that the exponents of the first terms (in bases $2$, $3$, etc., respectively) increase up to a certain base, say $b$, at which point we will have $(n)_{b-1}=(b)^{b}$. Next, we have $(n)_{b}=b\cdot(b+1)^{b}+\cdots+b\cdot(b+1)+b$, and the exponents never increase again as we continue the process. That is, for $k\ge b-1$ the exponents of the expression of $(n)_{k}$ are all $\leq b$. For example, take $n=7=2^{2}+2+1$. We have $$\begin{array}{c|c} b & (n)_{b-1}\\ \hline 2 & 2^{2}+2+1\\ 3 & 3^{3}+3\\ 4 & 4^{4}+3\\ 5 & 5^{5}+2\\ 6 & 6^{6}+1\\ 7 & 7^{7}\\ 8 & 8^{8}-1 = 7\cdot 8^{7}+\cdots+7\cdot 8+7\\ \vdots & \vdots\\ \end{array}$$
Thus, for $n=7$, the exponent reaches its maximum for the first time for base $b=7$ (i.e. for the $6^{th}$ term of the sequence). Up to $n=7$ the base $b$ is easy to find. But for $n=8$ it is a very large number, and things explode for larger values of $n$ (as one would expect).
How do we find this $b$ in general? Is there an explicit expression for $b$ perhaps in terms of the Hardy functions?