How we derive $y \geq (b+1)$ from $(b+1)b^n < (b+1)^n b$? In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page 5 that-
$y \geq (b+1)$ since, $(b+1)b^n < (b+1)^n b$ and it was given that  $(b+1)x^n - by^n =1$.
but $(b+1)x^n - by^n =1 \implies (b+1)x^n > by^n $, and together with   $(b+1)b^n < (b+1)^n b$, I don't see how we reach at $y \geq (b+1)$ 
Please explain, Thanks in advance.
 A: The context is that $b, x, y, n$ are positive integers here, and $(x, y) \neq (1, 1)$.
Under these conditions, $x < y$ must hold (neither $x = y$ nor $x > y$ can), thus $y \geqslant x + 1$, and therefore $(b + 1)x^n - b(x + 1)^n \geqslant 1$. So this is $> 0$, i.e.
$$\frac{b + 1}{b} > \Big(\frac{x + 1}{x}\Big)^n \geqslant \frac{x + 1}{x}$$
(and this even implies that $x > b$, i.e. $y > b + 1$).
A: The given equation is-
$(b+1)x^n - by^n =1 \cdots (1)$.
Case $1$:  If $x=y$, 
then equation (1) becomes $(b+1)x^n - bx^n =1 \implies bx^n+x^n - bx^n =1 \implies x^n =1 \implies x=1 $, but it is given that $(x, y) \neq (1, 1)$. So, $x \neq y$.
Case $2$:  If $x>y$,
then, let $x=y+c$ (where $c>0$, $c$ is an integer since x,y are integers) we substitute the value of $x$ in  equation (1) -
$(b+1)(y+c)^n - by^n =1\implies b(y+c)^n+(y+c)^n - by^n =1 $. By inspection, we find that the difference between $b(y+c)^n+(y+c)^n $ and $by^n$ is larger than $1$ (reader will see it more clearly if he expands $(y+c)^n $ (binomial expansion), note that $c$ is an integer).
Case $3$: So, $x<y$ (the only possibility). Let, $y=x+d$. Then equation (1) becomes-
$(b+1)x^n - b(x+d)^n =1 $ [substituting  the value of $y=x+d$ in  $(1)$]
$\implies \frac{b+1}{b} - (\frac{x+d}{x})^n =\frac{1}{bx^n} $ [dividing  both sides of the equitation by  $bx^n$ ]
$\implies \frac{b+1}{b} - (\frac{x+d}{x})^n >0 $ [Since, $\frac{1}{bx^n}>0$]
$\implies \frac{b+1}{b} > (\frac{x+d}{x})^n  $ 
$\implies \frac{b+1}{b} > \frac{x+d}{x}$ [Since, $(\frac{x+d}{x})^n > (\frac{x+d}{x}) $ ] 
$\implies 1+ \frac{1}{b} > 1 + \frac{d}{x} $
$\implies \frac{1}{b} >\frac{d}{x} $
$\implies x > bd $. Here, $d \geq 1$, we see that for $d=1, x>b$, as we have found earlier $y>x$, we deduce, $y>b \implies y\geq b+1$.
