Combination of Addition and Multiplication of 1 and 0 done 13^12 times to never reach above 3^3^3^3^3 I have been doing this question for some time.
If the question is asking us to find the combinations of adding and multiplying 1 and 0 why can't it be 1+1+1+1... but it can be 1+1=2 or just 1?
Below is the question:

Start with 0 and 1. Then do all combinations of adding and multiplying
2 of the numbres (with repetition) to get 0, 1, and 2. Repeat this
step 13¹² more times. Show that the smallest natural numbrer never
reached is greater than ⁵3 (3 raised to itself 5 times)

Any tips for how to get to the answer or what this question is asking us would be welcome! Thanks.
 A: I think I can try to clarify the question.
We begin with a set of $2$ numbers: $0$ and $1$. We then apply addition and multiplication to any two numbers from our set with repetition:
$$0 + 0 = 0 ;~ 0+1 = 1;~1+1=2;~0*0=0;~0*1=0;~1*1=1$$
By doing so we got $3$ numbers: $0$, $1$ and $2$. We now repeat the operation with the new set of numbers. After step $2$ we will get the set $\{0,1,2,3,4\}$. After step $3$ -- $\{0,1,2,3,4,5,6,7,8,9,12,16\}$, and this is the first example when we do not get just the set of the first consecutive natural numbers. (Which shows that the pattern is not that obvious.)
Now, consider we did $13^{12}$ such steps. The problem asks the following: show that all the numbers until ⁵3 will be represented in our set after all the steps. (The problem asks to show that the smallest number not in the set is greater than ⁵3 which is equivalent.)
Edit: Small hint for the solution: prove that all the numbers up to $2^{2^n}$ will be present after the $(n+2)$-nd step.
Edit 2: As Henry pointed out the first edit is incorrect.
