While messing around with the idea of ordinal collapsing functions, I stumbled upon an interesting simple function:

$$C(0)=\{0,1\}\\C(n+1)=C(n)\cup\{\gamma+\delta:\gamma,\delta\in C(n)\}\\\psi(n)=\min\{k\notin C(n),k>0\}$$

The explanation is simple. We start with $\{0,1\}$ and repeatedly add it's elements to themselves:

$C(1)=\{0,1,2\}\\ C(2)=\{0,1,2,3,4\}\\ C(3)=\{0,1,2,3,4,5,6,7,8\}\\\vdots\\C(n)=[0,2^n]$

And $\psi(n)$ is defined as the smallest integer not within $C(n)$, which is $2^n+1$.

I then extended my function. Imagine all the same definition, except that we now have

$$C(n+1)=C(n)\cup\{\gamma+\delta,\color{red}{\gamma\cdot\delta}:\gamma,\delta\in C(n)\}$$

This simple change gives us something a bit more complicated. The first few sets are

$C(1)=\{0,1,2\}\\ C(2)=\{0,1,2,3,4\}\\ C(3)=\{0,1,2,3,4,5,6,7,8,9,12,16\}\\ C(4)=\small\left\{\begin{align}0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 30,\\ 32, 35, 36, 40, 42, 45, 48, 49, 54, 56, 60, 63, 64, 72, 80, 81, 84, 96, 108, 112, 128, 144, 192, 256\end{align}\right\}\\ C(5)=\{0,\dots,177,179,\dots\}\\ \vdots$

If $\psi(n)$ is the smallest natural number not found in $C(n)$, asymptotically, how fast does $\psi$ grow?

The first few values of $\psi$ are

$$2,3,5,10,26,178,\dots$$

Here's a program that outputs $\psi$ and here's a program that outputs $C$.

I'm looking for better bounds and/or asymptotic formulas in the form of

$$\psi(n)\approx x^{y^n}$$