Inspired by this question, I'd like to know how one would go about proving the below more general equation?
$$n \in \mathbb{N},\;a \in \mathbb{N},\;b \in \mathbb{N}$$ $$b^n+1 \notin a^{\mathbb{N}+1}$$
$$\left\lfloor n\frac{\log (b)}{\log (a)}\right\rfloor =\left\lfloor \frac{\log \left(b^n+1\right)}{\log (a)}\right\rfloor$$
Empirically the above appears, prima facie, to be the case. If it's not, apologies, I tried quite a few values of $a$, $b$ and $n$ and came up with the above assumptions.
Even more is actually true for positive integers $x$ and $a>1$, such that $(x+1)$ is not a power of $a$:
$$\left\lfloor\frac{\log x}{\log a}\right\rfloor = \left\lfloor\frac{\log (x+1)}{\log a}\right\rfloor$$
In order to see why this is so, let $m$ be the largest power of $a$ which doesn't exceed $x$. This means that we have $$a^m \leq x < a^{m+1}$$ and also $$m \leq \log_a x < m + 1$$ Since $\log_a x = \frac{\log x}{\log a}$, we have $$\left\lfloor\frac{\log x}{\log a}\right\rfloor = m$$
The right-hand side floor expression is clearly not smaller than $m$, so we only need to prove it cannot be larger either. Assume (for a proof by contradiction) that this was the case to get $$\frac{\log(x+1)}{\log a} \geq \left\lfloor \frac{\log(x+1)}{\log a}\right\rfloor \geq m+1$$ In the light of inequality $x<a^{m+1}$, the only possibility would be $x+1=a^{m+1}$. But that's precisely what we've forbidden by the not-a-power condition, thus completing our proof.
