<s>[INST] Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$ 
Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$ where $n$ is a positive integer.

In order for a positive integer $k$ to begin with a $1$, we need $10^m \leq k < 2 \cdot 10^m$ for some positive integer $m$. Thus since $5^n$ begins with a $1$, we have $10^m \leq 5^n < 2 \cdot 10^m$ for some positive integer $m$. Then multiplying by $\dfrac{2^{n+1}}{5^n}$ we get $10^m \cdot \dfrac{2^{n+1}}{5^n} \leq 2^{n+1} < 10^m \cdot \dfrac{2^{n+2}}{5^n}$, but I didn't see how this helped.
 [/INST] Note that:
$$2\cdot10^n=5^n\cdot 2^{n+1}$$
If $5^n$ begins with $1$ then 
$$10^m<5^n<2\cdot10^m\to10^m< \frac{2\cdot10^n}{2^{n+1}}<2\cdot10^m$$
what give us
$$2^{n+1}< 2\cdot 10^{n-m}\text{ and } 2^{n+1}>10^{n-m}\Leftrightarrow 10^{n-m}<2^{n+1}<2\cdot 10^{n-m}$$
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