Growth rate of the nth natural number not constructable with n steps of addition and multiplication

Let's suppose that $\psi(n)=k+1$. It thus follows that

$$\{x:x\in[0,k]\}\subset C(n)$$

By adding these together, we may find that

$$\{k+x:x\in C(n)\}\subset C(n+1)$$

And thus, the simple bound of

$$\psi(n+1)\ge k+k+1=2\psi(n)-1$$

$$\psi(n)\ge2^n+1$$

By adding and multiplying, we find that

$$\{k+x,k\cdot x:x\in C(n)\}\subset C(n+1)$$

And by adding these again, we find that

$$\{k\cdot x_0+x_1:x_0,x_1\in C(n)\}\subset C(n+2)$$

Which encompasses all of the numbers from $0$ to $k^2+2k$, hence

$$\psi(n+2)\ge k^2+2k+1=\psi(n)^2$$

or,

$$\psi(n)\ge2^{2^{n/2}}$$

By adding and multiplying instead of just adding in the previous step, we find that

$$\{x_0,x_1k^2:x_0,x_1\in C(n+1)\}\subset C(n+2)$$

And by adding these, once again, we find that

$$\{x_1k^2+x_0:x_0,x_1\in C(n+1)\}\subset C(n+3)$$

which will encompass all of the numbers from $0$ to $2k^3+2k$, hence

$$\psi(n+3)\ge2k^3+2k+1=2\psi(n)^3-6\psi(n)^2+8\psi(n)-3$$

which is not at all pretty, but since $2\psi(n)\ge1$ for all $n$, we get

$$\psi(n+3)\ge2(\psi(n)-1)^3$$

And if $\psi(n)\ge\frac1{1-2^{-1/3}}\approx4.8$, then

$$2(\psi(n)-1)^3\ge\psi(n)^3$$

And thus,

$$\psi(n)\ge5^{3^{(n-2)/3}},{\rm~large~enough~}n$$

Doing this again gives

$$\psi(n+4)\ge2k^5+4k^4+2k^3+k^2+2k$$

And for large enough $k$,

$$2k^5+4k^4+2k^3+k^2+2k>(k+1)^5=\psi(n)^5$$

hence,

$$\psi(n)\ge5^{5^{(n-2)/4}},{\rm~large~enough~}n$$

and in general, I'm expecting

$$\psi(n)\ge2^{(2^x+1)^{(n-y)/(x+2)}}>2^{2^{x(n-y)/(x+2)}}$$

For all $x$, large enough $n$, and some $y$.

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