For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$? 
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?

This question comes from counting the number of digits of $10^n$ in terms of the number of digits of $2^n$ and $5^n$. Number of digits of $10^n$ is $n+1$ which equals to the  sum of the number of digits of $2^n$ and $5^n$. I know that the number of digits of a positive integer $x$ is $\lfloor \log_{10}(x)\rfloor + 1$.
Using programming, I've checked that this is true for $n$ less than $100$.

$\lfloor{\log_{10}(2^n)}\rfloor = \lfloor{\log_{10}(10^n) - \log_{10}(5^n)}\rfloor = n + \lfloor{ - \log_{10}(5^n)}\rfloor = n - \lfloor{\log_{10}(5^n)}\rfloor - 1$
So:
$\lfloor{\log_{10}(2^n)}\rfloor + \lfloor{\log_{10}(5^n)}\rfloor + 2 = n - \lfloor{\log_{10}(5^n)}\rfloor - 1 + \lfloor{\log_{10}(5^n)}\rfloor + 2 = n + 1$
