$2^n$ in base ten has two consecutive digits that are equal 
Determine (with proof) whether or not there exists a positive integer $N$ such that for every integer $n\ge N$, the number $2^n$ in base ten has two consecutive digits that are equal. (Trailing zeros do not count).

I tested many powers of $2$ for small $n$ and wasn't able to find such an $N$. Can we prove this by contradiction that there does not exist such an $N$? I was thinking that if $N$ does satisfy this, then we can consider certain powers of $2$ that will make a contradiction.
 A: I would say it is certain to be true, but will be difficult to prove.  Any number with thousands of digits is almost certain to have two consecutive digits equal.  We make the gross assumption that the digits are uniformly distributed and independent, which actually should be pretty good away from the ends of the number.  With a million digits, the chance that there is not a neighboring pair is $0.9^{1000000}\approx 3.23 \cdot 10^{-45758}$  As every $10$ powers of $2$ give three more factors $0.9$ this is essentially a geometric series with ratio $(0.9^3)^{1/10}=0.9^{0.3}\approx 0.96888$ and summing the series will multiply it by about $30$, leaving a vanishingly small number.
A: I'm not sure whether this will work, but looking to the last $k$ digits of $2^n$ might just work. 
When looking to the last two digits, we can see that for every $k \equiv 18 \mod 20$, we have $2^k \equiv 44 \mod 100$. Similiarly, if $k \equiv 19 \mod 20$, we have $2^k \equiv 88 \mod 100$.
Looking to the two digits before that, we need to use that the last three digits of $2^n$ repeat with period $100$. This solves the problem for all $k$ congruent to $3, 40, 41, 46, 53, 89, 90, 91, 96$, in addition to those that were already solved. 
There are now 19 congruence classes mod 100 eliminated. 
