How to prove $b^n-1 \geq n(b-1)$ for $b>1$ and $n \geq 0$ I have already figured out a simple proof by induction for this problem. Is there any other way to do it?

Prove $b^n-1 \geq n(b-1)$. For $b>1$ and $n \geq 0$. 

 A: Take $f(x) = x^n$, then mean value theorem says that there exists $c \in (1,b)$ such that $f(b) - f(1) = f'(c)(b-1)$, i.e. $$b^n - 1 = nc^{n-1}(b-1)$$
thus $b^n - 1 > n(b-1)$ since $c >1$. Actually we can see that one needs $n \geq 1$(or $n=0$) for the conclusion to be true
A: Actually the inequality is true for all $\color{red}{b>0}$.
The initialisation is trivial.
Inductive step:
Suppose $b^n-1>n(b-1)$  for some $n>1$. Rewrite this as $b^n>1+n(b-1)$ and multiply both sides by $b=b-1+1$, which is positive. We get:
\begin{align*}b^{n+1}&>1+b-1+n(b-1+1)(b-1)=1+(n+1)(b-1)+n(b-1)^2\\&>1+(n+1)(b-1).\end{align*}
A: MVT: Consider $f(x) = x^n \Rightarrow f'(x) = nx^{n-1} \geq n, x \geq 1 \Rightarrow f(b) - f(1) = f'(c)(b-1) \Rightarrow b^n-1 \geq n(b-1)$
Bernoulli: Take $b = 1+a, a > 0 \Rightarrow b^n - 1 = (1+a)^n - 1 \geq 1+an-1 = an = n(b-1)$.
A: $$b^n-1=(b-1)(b^{n-1}+\cdots+b+1)\ge(b-1)(1+\cdots+1)=n(b-1).$$
A: One way: $b^n-1=(b-1)(1+b+...+b^{n-1})>(b-1)()1+1+...+1) =n(b-1)$
Other way: consider function $f(b)=b^n-1-n(b-1)$, for $b>1$. Taking derivative and conculde that it's increasing on $b \in (1,\infty)$. Thus $f(b)>f(0)=0$.
A: Yes, there is another answer: we have that 
$$(b - 1) ( b^{n - 1} + b^{n - 2} + \cdots + b + 1) = b^{n} - 1.$$
This means that we can write 
$$ \frac{b^n - 1}{b - 1} = b^{n - 1} + b^{n - 2} + \cdots + b + 1.$$
The right hand side has $n$ terms all of which (by hypothesis on $b$) are greater than or equal to 1; thus we can conclude that 
$$ \frac{b^n - 1}{b - 1} \ge n.$$
A: Here is yet another way...
Case 1:  $b, n \ge 1$
$$x > 1\implies x^{n-1} \ge 1 \implies \int_1^b x^{n-1} dx \ge \int_1^b 1 dx \implies b^n-1 \ge n(b-1)$$
Case 2:  $b \in (0, 1)$ and $n \ge 1$.
$$x < 1\implies x^{n-1} \le 1 \implies \int_b^1 x^{n-1} dx \le \int_b^1 1 dx \implies b^n-1 \ge n(b-1)$$
Note however that if $n \in (0, 1)$ your inequality does not hold, rather the reverse does.  Same proof...
A: I had also noticed @PetiteEtincelle's counterexample, $(b,n) = (4,\frac{1}{2})$.  And this suggested an intended fixup and proof...
The stated inequality holds for $b,n \in [1,\infty) \subset \mathbb{R}$.
True by inspection at $b=1,n=1$.  In fact, true by inspection at $b=1, n \in [1,\infty)$.  Partial differentiation with respect to $b$ gives $n b^{n-1} \geq n$ which is evidently strictly true for $b>1, n>1$.  So for any point $(b,n) \in [1,\infty) \times [1,\infty)$, we can start at $(1,n)$, where the stated inequality holds, then increase the first coordinate to $b$, and the inequality will hold at every point along the path (because our derivative shows the "big side" grows strictly faster along this path than the "small side").
BTW:  Before anyone "complains"...  Yes, I know that the notation "$n$" suggests that $n \in \mathbb{Z}$ or $\mathbb{N}$.  But "$b$" doesn't suggest anything, so using "obvious" default domains isn't so obvious.
A: Proof By Induction: 
Inductive Hypothesis: $b^{n}-1\geq n(b-1)$
Base: $n=0$. $0=0$ True.
 Inductive Step:
$b^{n+1}-1\geq (n+1)(b-1)$
$b^{n+1}-1\geq n(b-1)+b-1$
$b^{n+1}\geq n(b-1)+b$
$b*b^{n}\geq b*[n(b-1)+1]$$b*[n(b-1)+1]\geq n(b-1)+b$ $\implies b\geq 1$, Hypothesis. QED
