prove $\exists N,\forall n\ge N$, there are more powers of $2$ within $2^1,2^2,\cdots, 2^n$ that start with $7$ than $8$. 
Prove that there exists an integer $N$ so that for any $n\ge N,$ there are more powers of $2$ within $2^1,2^2,\cdots, 2^n$ that start with $7$ than $8$.

Note that there are infinitely many powers of 2 starting with any given positive integer. Also that $0.8 \cdot 0.875 = 0.7, 0.9\times 0.876 = 0.7884.$ The statement that a power of $2$, say $2^m$, starts with $7$ is equivalent to saying that there exists a power of $10$, say $10^k$ so that $7\cdot 10^k \leq 2^m < 8\cdot 10^k,$ or that $\dfrac{2^m}{10^k}\in [7,8)$ or equivalently that $m' - k\log_2 5 \in [\log_2 7, 3)$ for $m'=m-k$. It might be useful to try a few specific cases of powers of two to get an idea of how to select $N$. Note that by this link the first power of $2$ starting with 7 is $2^{46}$ and the first one starting with $9$ is $2^{64}$. Or it might be easier to use an existence proof similar to the group-theoretic proof that there are infinitely many powers of 2 starting with any given positive integer.

A digit $d$ is the starting digit of $2^n$ if and only if there exists some $m$ for which
$$d10^m\leq 2^n<(d+1)10^m.$$
This $m$ is exactly $\lfloor n\log_{10}2\rfloor$, and so one has
$$d=\left\lfloor\frac{2^n}{10^m}\right\rfloor=\left\lfloor\frac{2^n}{10^{\lfloor n\log_{10}2\rfloor}}\right\rfloor=\left\lfloor 10^{\{n\log_{10}2\}}\right\rfloor,$$
where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$. Put another way, $d$ is the starting digit of $2^n$ if and only if
$$\log_{10}d\leq \{n\log_{10}2\}<\log_{10}(d+1).$$
Now, note that $\log_{10}2$ is irrational, so the sequence $a_n=\{n\log_{10}2\}$ is uniformly distributed within $[0,1)$. That is, as $N\to\infty$, for any $0\leq a<b<1$,
$$\frac{\text{# of }1\leq n\leq N\text{ with }a\leq \{n\log_{10}2\}<b}{N}$$
tends to $b-a$. In particular, the proportion of $n$ for which $2^n$ has first digit $7$ tends to $\log_{10}(8/7)$, while the proportion of $n$ for which $2^n$ has first digit $8$ tends to $\log_{10}(9/8)$. Since $\log_{10}(9/8)<\log_{10}(8/7)$, eventually the first quantity is larger than the second, as desired.
