If we define $$\cal N _1 := \{ 1\} $$ and by induction $$\cal N_{n+1}:=\{x\in \mathbb N | \exists a,b \in\cal N_n : x= a+b \text{ or }x=ab \text{ or }x=a^b \}$$
it's easy to prove that, for every $m \in \mathbb N\backslash \{0\}$, there exists $n\le m$ such that $m\in \cal N_n$.

If we define $\deg (n):= \min\{m\in \mathbb N | \;n\in\cal N_m\}  $ (let's call it degree of n), can you find a way to calculate the degree of every positive integer (which is at least better than brute force)?  

Not only is it possible to find a multiple of $n$ whose decimal expansion consists solely of $0$s and $1$s, it is possible to arrange for all the $1$s to come before all the $0$s.

Suppose first that $n$ is coprime to $10$. Then by Fermat–Euler, $10^{\varphi (9n)} \equiv 1 \pmod{9n}$. Thus $(10^{\varphi (9n)} -1)/9 \equiv 0 \pmod{n}$, and so there is a multiple of $n$ which consists solely of $1$s, namely $(10^{\varphi (9n)} -1)/9$.

Now consider the opposite case that $n=2^a 5^b$ for some natural $a, b$. Then some multiple of $n$ is a power of $10$: either $2^{b-a}n$ or $5^{a-b}n$, depending on whether $a$ or $b$ is greater.

Thus for general $n$, we can express $n$ as $2^a 5^b m$, where $m$ is coprime to $10$. Then we can find a multiple of $m$ which is a string of $1$s and a multiple of $2^a 5^b$ which is a power of $10$, and hence a multiple of $n$ which is a string of $1$s followed by a string of $0$s. Specifically, there are $\varphi (9m)$ $1$s and $\max(a,b)$ $0$s, so $\varphi (9m)+\max(a,b)$ gives an upper bound on the number of digits needed.
