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I was thinking about this today. $2^6$ is easy, $64$ and then $2^{10}$ is roughly $1000$ so $2^{16}$ is really $(2^6)(2^{10})$ or $(2^6)\cdot 1024$ and this nicely gives a result that starts with $6$ up to $2^{46}$. Beyond that the numbers are so large that they seem to all contain a $6$ as part of their digits.

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    $\begingroup$ Not an answer, but note that $2^{6+10x}=4^{3+5x}$. And $4^n$ has a $6$ in the single digits' place for all even $n\in\mathbb{N}$, and a $4$ for all odd $n$. And so whenever $3+5x$ is even (i.e. when $x$ is odd), a six will be present. $\endgroup$ – Brent Mar 21 '15 at 2:46
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The last two digits always alternate between 64 and 36 which answers the question positively.

Note that $2^{10}=1024$ , the last two digits being $24$.

Note that $36*24=864$ , the last two digits are $64$.

And $64*24=1536$ , the last two digits are $36$.

So, if we restrict our attention to only the last two digits, every time
we multiply by $2^{10}=1024$ it is like multiplying by $24$.
If the last two digits were $64$ then they change to $36$.
If the last two digits were $36$ then they change to $64$.
So the last two digits always alternate from $64$ to $36$, and from $36$ back to $64$.
In particular, $6$ does always appear as one of the last two digits.

Just to illustrate the above explicitly:
$2^6=64$
$2^{16}=65536$
$2^{26}=67108864$
$2^{36}=68719476736$
$2^{46}=70368744177664$
$2^{56}=72057594037927936$
etc.

Motivated by the above question (and its answer), I posted a follow up.

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