Counterintuitive Goodstein's Theorem There is an amazing and  counterintuitive  theorem: 

For all $n$, there exists a $k$ such that the $k$-th term of the Goodstein sequence $G_k(n)=0$. In other words, every Goodstein sequence converges to $0$.

How can I find $N$ such $G_{N}(n)=0$?
for instance if $n=100$
$G_0(100)=2^{2^{2}+2}+2^{2^{2}+1}+2^2=100$
$G_1(100)=3^{3^{3}+3}+3^{3^{3}+1}+3^3-1=228767924549636$
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How to find $k$ such $G_{k}(100)=0$?
 A: You don't. 
Let $g$ be the "Goodstein terminating" function: $g(n)=k$ iff $G_k(n)=0$ and $G_{k-1}(n)\not=0$. Then $g$ grows insanely fast: if you've heard of the Ackermann function, it's of a similar species. In fact, new notation and concepts had to be invented to even talk about how fast such functions grow! Look up "fast-growing hierarchies" (or see https://en.wikipedia.org/wiki/Goodstein%27s_theorem#Sequence_length_as_a_function_of_the_starting_value).
In fact, $g$ grows so fast, that when I say $$\mbox{$g(100)$ has more digits than there are atoms in the known universe,}$$ that's an understatement.
A: We happen to have an approximation to this sequence using a thing called the Hardy Hierarchy.  Here, we happen to have

$$k_{100}\approx H_{\omega^{\omega^\omega+\omega}+\omega^{\omega^\omega+1}}(2^{402653184}402653184-1)$$

This may be noted by observing the behavior of the Goodstein sequence:
$$n^ka_k+n^{k-1}a_{k-1}+\dots+na_1+a_0$$
If $a_0>0$, then this behaves like a successor ordinal:
$$n^ka_k+n^{k-1}a_{k-1}+\dots+na_1+a_0\\\to(n+1)^ka_k+(n+1)^{k-1}a_{k-1}+\dots+(n+1)a_1+a_0-1$$
The Hardy Hierarchy works similarly:
$$H_{\alpha+1}(n)=H_\alpha(n+1)$$
As you can see, the base in both goes up by one, while the overall number goes down by one.
If $a_p=0$ for $p\in[0,j]$, then this behaves similarly to the limit ordinal:
$$n^ka_k+n^{k-1}a_{k-1}+\dots+n^{j+1}a_j\\\small\to(n+1)^ka_k+(n+1)^{k-1}a_{k-1}+\dots+(n+1)^{j+1}(a_j-1)+(n+1)^jn+\dots+(n+1)n+n$$
For the Hardy Hierarchy:
$$H_{\omega^ka_k+\omega^{k-1}a_{k-1}+\dots+\omega^ja_j}(n)=H_{\omega^ka_k+\omega^{k-1}a_{k-1}+\dots+\omega^j(a_j-1)+\omega^j(n-1)+\dots+\omega(n-1)+n}(n)$$
This is the same with the very small exception that the base (the number in the parenthesis) does not go up.  However, this is extremely close to the Goodstein sequence...
So this is basically how I got the approximation, after expanding out $g_{2^{402653184}402653184-1}(100)$ (easily done once you've spotted the patterns)
All that being said, this approximation only goes to show that this is much larger than what most people could fathom.  In terms of the fast growing hierarchy, which a few people may know better, we have
$$k_{100}\approx f_{\omega^\omega+\omega}(f_{\omega^\omega+1}(2^{402653184}402653184-1)$$

As Deedlit notes, the base being off can be fixed by considering
$$H_{\omega^ka_k+\omega^{k-1}a_{k-1}+\dots+\omega^ja_j}(n+1)$$
Thus, we actually get an exact value:
$$\begin{align}k_{100}&=H_{\omega^{\omega^\omega+\omega}+\omega^{\omega^\omega+1}+\omega^\omega}(3)-3\\&=H_{\omega^{\omega^\omega+\omega}}(H_{\omega^{\omega^\omega+1}}(H_{\omega^\omega}(3)))\\&=f_{\omega^\omega+\omega}\left(f_{\omega^\omega+1}\left(2^{2^{24}\cdot24}\cdot2^{24}\cdot24\right)\right)\end{align}$$
