Multiplying $1$ by $5^a$ is the same as multiplying by $10^a$ and dividing by $2^a$. In other words, $5^a$ has the same decimal representation as $2^a$ once we remove the decimal point (and the leading 0s). Then the problem can be re-framed as showing that any consecutive sequence of leading digits in the decimal representations of fractions $\frac{1}{2^a}$ also appears among consecutive leading digits of decimal representations of powers $2^b$, $b \in \mathbb{N}$ (not reversed). Let $f(x)$ denote the leading digit of $x \in \mathbb{R}^{+}$ (where, as above, the leading digit of $0.a_1 a_2 a_3 \ldots$ is $a_1$ if $a_1 \neq 0$, otherwise it is the first non-zero digit beyond the decimal point). Then it suffices to show that any finite sequence of consecutive elements of $\ldots, f(2^{-2}), f(2^{-1}), f(2^0), f(2^1), f(2^2), \ldots$ occurs again farther to the right in the sequence. To do this, we find powers of $2$ that are very close to the next smallest power of $10$. Fix $N \ge 0$ and let $M$ be such that $\forall r \le N$, $5^r < 10^M$. Choose $a \in \mathbb{N}$ such that: \[0 < a \log_{10} 2 - \lfloor a \log_{10} 2 \rfloor < \log_{10}\left(1 + \frac{1}{10^M}\right)\]This is achievable since $\{\log_{10} 2\}$ is irrational, so that a short pigeonhole principle argument proves there is such $a$. Let $g = \lfloor a \log_{10}2 \rfloor$. For $m \le N$, $5^m = c \cdot 10^d + b$ for $b < 10^d$ and $c \in \{1,2,\ldots, 9\}$. Then \begin{align*} &c \cdot 10^d + b < 10^{d+1} \le 10^M \\ \Longrightarrow & \frac{1}{10^M} < \frac{1}{c \cdot 10^d + b} \le \frac{10^d - b}{c \cdot 10^d + b} \\ \Longrightarrow & 1 + \frac{1}{10^M} < \frac{10^d(c+1)}{c \cdot 10^d + b} \\ \Longrightarrow & (c \cdot 10^d + b)\left(1 + \frac{1}{10^M}\right) < 10^d (c+1) \\ \Longrightarrow & 5^m 10^g \left(1 + \frac{1}{10^M}\right) < 10^{d + g}(c+1) \\ \Longrightarrow& 10^{d+g}c < 5^m 2^a < 10^{d+g}(c+1) \end{align*} So for all $r \le n$, $f(5^r) = f(5^r \cdot 2^a)$ (so that also $f(2^{r}) = f(2^{r} \cdot 2^a)$. Then for any subsequence whose left most term is at least $2^{-\lfloor N / 2 \rfloor}$, keep multiplying by $2^a$ until the entire sequence has positive exponents. Throughout the operation $f(2^r) = f(2^r 2^a) = f(2^r 2^{2a}) = \cdots$, so the claim follows.