Maybe this is wrong, but I'll give it a try... I will use a lemma. Lemma: Let $N$ be a given natural number and $A=\overline{a_i....a_0}$. Then there exist infinitely many $k_0\in\mathbb N$ such that the decimal representation of $2^{k_0N}$ starts with the digits of $A$. Proof: The lemma is slightly different than this http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1875787#p1875787 but the proof is exactly the same. Back to the original problem. I will show that the given number is irrational. To prove this, it is enough to show that the sequence $a_n$ is not periodic. For the sake of contradition, I assume that it is periodic with period $N$. Then $\displaystyle a_{kN+u}=a_u\forall \ k\in\mathbb N_0, \ 1\leq u\leq N$. Taking $u=N$, I have that the leftmost digit of $2^{kN}$ is constant for all $k\geq1$. But this is wrong because from the lemma I may choose a $k_0$ which gives another leftmost digit. This means that my assumption was wrong and $a_n$ is not periodic.

I think that the number is rational, cause it is 10-periodic.. But I don't know how to prove my asertion.

@vlad1m1r: $2^{320}$ begins with a $2$ yet $2^{10}$ begins with a $1$. The sequence is not $10$-periodic.

Isn't there some easier way ? I was trying to derive a contradiction but the problem boils down to the lemma proposed by @Mikesar Isn't there some easier way to prove the lemma ???

I think, You couldn't avoid using some kind of Kronecker's density theorem or equidistribution of the irrational rotation. Here is another way of doing it. The rate how often the digit $j\,,\, 1\le j\le 9$ occurres in the sequence $a_1,a_2,\ldots$ is positive and different for every digit. We can explicitly calculate this rate. I have recently met this result as an interesting application of the equidistribution of the sequence $\{n\alpha\},n=1,2,\ldots$, where $\alpha $ is an irrational, $\{x\}$ is the fractional part of $x$. As a bonus we get the irrationality of the given number, since otherwise the digits taking part in the period will have positive rational rates of occurrence and the others' oocurrence rate will be zero. As could be seen below, it's not the case. The first digit of $2^n$ is $j$ iff $ j 10^m \le 2^n < (j+1)10^{m}$ for some $m\in \mathbb{N}$. It means $\{n\lg 2\} \in [\, \lg j , \lg(j+1) \,)$. \[ |\{n\mid n\le N, j\,\, \text{is the first digit of }\,\, 2^n\}|/N = \frac{1}{N}\sum_{i=1}^N \chi_{[\,\lg j , \lg(j+1)\, )}\, \{i\lg 2\}\] Due to the equidistribution of $\{n\alpha\}$, when $\alpha$ is irrational, the RHS tends to $\lg\frac{ j+1}{j}$, when $N\to \infty$, so it is the rate of occurrence of the digit $j$. Apparently $1$ occurres most frequently than the other digits.

yeah, just use the fact that the sequence of a_i should has a term. a_n*10^b_n<2^n<(a_n+1)10^b_n a_n*10^b_(n+k)<2^(n+k)<(a_n+1)10^b_(n+k) ... that means that a_n/(a_n+1)*10^(b_(n+k)-b_n)<2^k<(a_n+1)/(a_n))*10^(b_(n+k)-b_n) and it must be also true at nk. that will be false because it squeeze into 10^t but 2^k could not squeeze in