Can you prove or disprove this floor/ceiling equation for positive integers $\displaystyle k$ and $\displaystyle b$, $\displaystyle b \geq 2$?

$\displaystyle \lfloor \log_{b}{k} \rfloor + 1 = \lceil \log_{b}{\left( k+1 \right)} \rceil$

I think I can prove the equation when either $\displaystyle k$ or $\displaystyle k+1$ is a positive integer power of $\displaystyle b$ as follows.

When $\displaystyle k=b^n$ for positive integer $\displaystyle n$, the original equation becomes $\displaystyle \lceil \log_{b}{\left( b^n + 1 \right)} \rceil = n + 1$. Which can be proven because $\displaystyle 0 < \log_{b}{\left( b^n + 1 \right)} - n < 1$.

Similarly, when $\displaystyle k+1=b^n$ for positive integer $\displaystyle n$, the original equation becomes $\displaystyle \lfloor \log_{b}{\left( b^n-1 \right)}\rfloor = n-1$. Like before, this is easy to prove because $\displaystyle 0 < n - \log_{b}{\left( b^n-1 \right)} \leq 1$.

Is this correct so far? I'm not sure what implications this has for when neither $\displaystyle k$ nor $\displaystyle k+1$ is a positive integer power of $\displaystyle b$. Any pointers you can offer will be greatly appreciated. Thanks!