Show that these two numbers have the same number of digits

I want to show that for $n>0$, ${2}^{n}$ and ${2}^{n}+1$ have the same number of digits.

What I did was I found that the formula for the number of digits of a number $x$ is $\lfloor {\mathrm{log}}_{10}(x)\rfloor +1$, so basically if I subtract that formula with $x={2}^{n}$ with the formula with $x={2}^{n}+1$, I should get zero.

$\lfloor {\mathrm{log}}_{10}({2}^{n})\rfloor +1-(\lfloor {\mathrm{log}}_{10}({2}^{n}+1)\rfloor +1)=\lfloor {\mathrm{log}}_{10}({2}^{n})\rfloor -\lfloor {\mathrm{log}}_{10}({2}^{n}+1)\rfloor $

At this point, I don't know of a way to simplify this any further to make it equal $0$. I thought about mentioning that ${\mathrm{log}}_{10}(x)$ increases slower than $x$ as $x$ increases, which would mean the difference of the floor of the logs of two consecutive numbers may be close to zero, but that doesn't cut it to prove that ${2}^{n},{2}^{n}+1$ have exactly the same number of digits.

Are there any special floor or log properties I could use to make this easier? Any help is appreciated.