Infinitely many $n$ such that $2^n$ in base $10$ starts with $7777777$

As noted by @Winther, $2^n$ starts with $777777$ if and only if $777777\cdot 10^m \le 2^n < 777778\cdot 10^{m}$ for some $m \in \mathbb{N}$.

This can be rewritten as

$$m + \log_{10} 777777 \le n\log_{10}2 < m + \log_{10} 777778$$

Let $\alpha = \log_{10}2$ and conclude there are infinitely many $n \in \mathbb{N}$ such that $$n\log_{10}2- \left\lfloor n\log_{10}2\right\rfloor \in \big[\log_{10} 777777 - 5, \log_{10} 777778-5\big\rangle \subseteq [0,1]$$

For these $n$ we have

$$(\left\lfloor n\log_{10}2\right\rfloor - 5) + \lfloor n\log_{10}2\rfloor + \log_{10} 777777 \le n\log_{10}2 < (\left\lfloor n\log_{10}2\right\rfloor - 5) + \log_{10} 777778$$

so for them $2^n$ starts with $777777$.

The key fact is this (see here for a proof):

For every finite sequence of decimal digits, there is a power of $2$ whose decimal expansion begins with this sequence.

Now apply this to $7777777\cdots 7$, with an increasing number of $7$s.