No need to use big guns here ... just focus on the last digit as you keep multiplying:
8,4,2,6,8,...
OK, so it cycles with a period of 4
Since 2011 divided by 4 leaves 3, we should be looking at the third entry of this cycle, i.e. it's a 2
Making a new answer since my first answer was from a misunderstanding of the question.
We can compute the problem modulo any reasonably small $M$ using a recursive algorithm as follows. I'll use the (in my opinion) simpler notation $m_n$ for what you've defined as $min\_number(x)$, and I'll define $a_n = (m_n-1)/9$. The goal will be to compute $a_n$ modulo $M$, since $m_n$ can be trivially recovered from $a_n$.
First note that using the Chinese remainder theorem, we can always assume that $M = p^\ell$ is a prime power.
Now first suppose $p\neq 2,3,5$. Then we can recursively compute $a_{n-1}\mod \phi(M)$, and since $9$ is invertible mod $M$ we can compute $a_n\equiv 9^{-1}\cdot (2\cdot10^{a_{n-1}\mod\phi(M)}-2)\mod M$.
Now suppose $p=2$ or $p=5$. I'll make the basic assumption that $\ell < (m_4-1)/9$. Note that $\ell$ is basically the log of $M$, so this is hardly an assumption. But looking at the recursive formula, clearly if $M=2^\ell$ or $M=5^\ell$ with $\ell < (m_4-1)/9$, then $m_n \equiv -1\mod M$ for all $n>4$. Again since we have a modular inverse for $9$ mod $M$, we can compute $a_n \equiv -2\cdot 9^{-1}\mod M$ for $n>4$.
Finally we consider $p=3$. Now to compute $a_n$ modulo $3^\ell$, it suffices to compute $m_n$ modulo $3^{\ell+2}$; indeed, suppose $3^{\ell+2} + k = m_n = 9a_n+1$. Then subtracting one from both sides and dividing by $9$, we get $3^{\ell} + (k-1)/9 = a_n$, i.e. $a_n \equiv (k-1)/9 \mod 3^{\ell}$ iff $m_n\equiv k \mod 3^{\ell+2}$.
Now we can recursively compute $a_{n-1}\mod \phi\left(3^{\ell+2}\right)$, and derive $m_n\equiv 2\cdot10^{a_{n-1}\mod\phi\left(3^{\ell+2}\right)}-1\mod 3^{\ell+2}$, and we're done.
Note that even though the CRT step branches the process, it only branches it into prime powers smaller than $M$. Thus if you simultaneously compute the values of $a_n$ for every prime power smaller than $M$, the complexity is actually bounded in time $O\left(\frac{M}{log(M)}n\right)$ (note that $\phi(3^\ell) = 2\cdot 3^{\ell-1}$, so even though $M$ gets bigger when it's a power of $3$, this only increases the number of values we need to compute by a term linear in $n$).
Best Answer
It's a bit unclear to me what you mean by "problem where there are an N amount of power values in the range of 1-9 and the last number has to be found". Is this to say you're dealing with a power tower of integers in the range 1-9?
In any case, to find the last digit of $9^{4^{2^{3^5}}}$, we want to "reduce it mod $10$". First, some notation. \begin{equation*} a \equiv b \pmod{10} \end{equation*} means that "$a$ is congruent to $b$ modulo $10$". This means that they differ by a multiple of $10$, or equivalently, they leave the same remainder when divided by $10$. Hopefully you agree that if some integer \begin{equation*} n \equiv 3 \pmod{10} \end{equation*} then the last digit of $n$ must be $3$.
Now, we also have that if $a \equiv b \pmod{10}$ and $c \equiv d \pmod{10}$, then $ac \equiv bd \pmod{10}$. Maybe this is obvious to you already, if not, you can prove it from the "differ by a multiple of $10$" definition. Using this, any power of $9$ is congruent to $-1$ raised to that same exponent. So \begin{equation*} 9^{4^{2^{3^5}}} \equiv (-1)^{4^{2^{3^5}}} \pmod{10} \end{equation*} But the value of $(-1)^{4^{2^{3^5}}}$ just depends on whether $4^{2^{3^5}}$ is even or odd. Since this is some power of $4$, which is even, it is clearly even (since it is not $4^0$), so $(-1)^{4^{2^{3^5}}} = 1$, so the last digit of $9^{4^{2^{3^5}}}$ is $1$.
An alternative approach would just be to look at powers of $9$ mod $10$, and notice that they are congruent to $1$, $9$, $1$, $9$, ... and take it from there.
To generalise this approach for some kind of program requires looking at the periodicity of the powers of the base modulo $10$, and then recursing into the power tower to reduce the subtower modulo the period of the powers of the base. (Particularly, not all numbers are as nice as $9$, which only has period $2$)
Proof of $a \equiv b \pmod{10}$ and $c \equiv d \pmod{10} \implies ac \equiv bd \pmod{10}$.
Have $a = b + 10n$, $c = d + 10m$ for some $n, m \in \mathbb Z$ by definition.. Then \begin{align*} ac &= (b + 10n)(d + 10n) \\ &= bd + 10(n + m + 10nm) \\ &\equiv bd \pmod{10} \quad \text{by definition} \end{align*}
Proof of $a \equiv b \pmod{10} \implies a^n \equiv b^n \pmod{10}$:
The base case $n = 0$ is trivially true. Then \begin{align*} a^{n + 1} &\equiv a \cdot a^n \\ &\equiv b \cdot a^n \quad \text{by assumption} \\ &\equiv b \cdot b^n \quad \text{by inductive hypothesis} \\ &\equiv b^{n + 1} \end{align*}