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.
 A: First, note that the inequalities must be strict, as we cannot have $5^n = 10^m$ for natural numbers $n,m$. Since $10^m = 2^m5^m$, your last equation becomes
$$2^{n+m+1}5^{m-n} < 2^{n+1} < 2^{n+m+2}5^{m-n}$$
$$\implies 2^{1+2n}10^{m-n} < 2^{n+1} < 2\cdot 2^{1+2n}10^{m-n}$$
Dividing each term by $2^{1+2n}$ gives us
$$10^{m-n} < 2^{-n} < 2\cdot 10^{m-n} $$
Taking the reciprocal of each term,
$$10^{n-m} > 2^n > \frac 1 2 10^{n-m} $$
Finally, multiplying by $2$ gives us
$$2\cdot 10^{n-m} > 2^{n+1} > 10^{n-m}$$
A: 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
$$2^{n+1}< 2\cdot 10^{n-m}\text{ and } 2^{n+1}>10^{n-m}\Leftrightarrow 10^{n-m}<2^{n+1}<2\cdot 10^{n-m}$$
A: $10^m < 5^n < 2*10^m$ [Note: equality can not hold as $5^n$ has no factor of $2$.]
$10^m < \frac {10^n}{2^n} < 2*10^m$
$1 < \frac {10^{n-m}}{2^n} < 2$
$2^n <  10^{n-m} < 2^{n+ 1}$
So $10^{k} < 2^{n+1}$ for $k = n-m$.
...
And $2^m5^m < 5^n < 2^{m+1} 5^m$
$2^{n+1}5^m < 2^{n-m + 1}5^n$
$2^{n+ 1} < 2^{n-m+1}5^{n-m}$
$2^{n+1} < 2*10^{n-m}$
So $2^{n+1} < 2*10^k$ for $k = n-m$
So $10^k < 2^{n+1}< 2* 10^k$.
So $2^{n+1}$ has a first digit of $1$.
