Questions on proofs that a number's digits = $\lfloor \log_{10} n \rfloor +1$?

Clearly, for a natural number $d$, the smallest $d$-digit natural number is $10^{d-1}$ -- a one followed by $d-1$ zeroes. Likewise, the smallest $(d+1)$-digit number is $10^d$, hence it follows that the largest $d$-digit number is $10^d-1$. Hence $n$ has $d$ digits if and only if $$ 10^{d-1}\le n < 10^d.$$ By taking logarithms: if and only if $$ d-1\le \log_{10}n<d.$$ In other words:

$d-1$ is the greatest integer less than or equal $\log_{10}n$.

But "the greatest integer less than or equal" is precisely what the floor function $\lfloor \cdot\rfloor$ denotes. So $$ d-1=\lfloor\log_{10}n\rfloor.$$ Or: $$ d=\lfloor\log_{10}n\rfloor+1.$$