A fun problem by Arnold using the Poincaré recurrence theorem

$2^{46}=70368744177664$ so 7 does definitely appear.

The first digit of a number $x$ can be found by:

$$\left\lfloor10^{frac\left(\log x\right)}\right\rfloor$$

Here $\log$ is the base 10 logarithm. Thus the first digit is 7 iff:

$$\log7 \le frac\left(\log x\right) \lt \log8$$

And it is 8 iff:

$$\log8 \le frac\left(\log x\right) \lt \log9$$ Now since

$$\log{2^n}=n\log2$$

We get that any range of values within $\left[0;1\right)$ will be represented, and their frequency will be proportional to the size of the range (ie. the distribution is uniform). Since $\log9-\log8\lt\log8-\log7$, we should see the digit $7$ more often than $8$, and in general, the smaller the digit, the more often it will appear.

Indeed, I made my computer calculate the first digits of $2^n$ for $0\le n\lt 1000000$, and it gave me: $$ \begin{matrix} 1: & 301030 \\ 2: & 176093 \\ 3: & 124937 \\ 4: & 96911 \\ 5: & 79182 \\ 6: & 66947 \\ 7: & 57990 \\ 8: & 51154 \\ 9: & 45756 \\ \end{matrix} $$

You will notice that these counts are all roughly equal to $1000000\cdot\left(\log{\left(d+1\right)}-\log{d}\right)$

A hint:

One has $x_n:=\log_{10}\bigl(2^n\bigr)=n\>\kappa$, where $\kappa:=\log_{10}(2)$ is irrational. It follows that the $x_n$ are uniformly distributed modulo $1$. See also Benford's law.

You don't need Poincaré's recurrence theorem for this.