# Prove $x^n -1 \ge (x-1)^n \ \forall n \in \mathbb{N}\ \land \ \forall x\ \ge 1$

Does my proof make sense?

Problem: prove $$x^n -1 \ge (x-1)^n \ \forall n \in \mathbb{N}\ \land \ \forall x\ \ge 1$$

Base step (P(1)):$$\ \ \ \ \ \ \ \ x-1 \ge x-1$$

Hypotesis (P(n)): $$\ \ \ \ \ x^n -1 \ge (x-1)^n$$

Thesis (P(n+1)):$$\ \ \ \ \ \ \ \ x^{n+1}-1 \ge (x-1)^{n+1}$$

Induction step: $$x^{n+1}-1 = x^n \cdot x -1 = x \cdot x^n -1 \ge x \cdot (x-1)^n \ge (x-1)^{n+1}$$ where: $$x \cdot (\not x\not-\not1\not)^n \ge \not(\not x\not-\not1\not)^n \cdot (x-1)$$ so $$x \ge x-1$$

$$x^{n+1}-1$$ $$=(x-1+1)x^n-1$$ $$=(x-1)x^n + x^n-1$$ $$\ge (x-1)x^n + (x-1)^n$$ $$\ge (x-1)x^n$$ $$\ge (x-1)^{n+1}$$

Your solution isn't valid as $$x\cdot x^n-1\ne x(x^n-1)\;\;\forall x\in\mathbb{R}$$.

• Very clear answer! just one doubt: how did you get to know to write $(x-1+1)x^n$ just because we must lead us to $(x-1)x^n$? Apr 2 '21 at 14:07
• Basically because we want to get from $(x-1)^n$ to $(x-1)^{n+1}$ and this technique looks like a good start. A possible next step could be $(x-1)x^n\ge (x-1)((x-1)^n+1)$.
– JMP
Apr 2 '21 at 14:10
• Ok thanks!, can I know if my solution is valid as well or not please? Apr 2 '21 at 14:12

Well if you have $$x^n-1\ge (x-1)^n$$ then multiply it by $$(x-1)$$ to get: $$x^{n+1}+1-x-x^n=(x^n-1)(x-1)\ge (x-1)^{n+1}$$ and the LHS is less than $$x^{n+1}-1$$.

$$(x-1)^n=\binom{n}{0}(-1)^n+\binom{n}{1}(-1)^{n-1}x+\cdots+\binom{n}{n}x^n$$ by the Binomial Theorem.

Evaluating the above at $$x=1$$ we see

$$\sum\limits_{i=0}^n(-1)^{n-i}\binom{n}{i}=0, \sum\limits_{i=0}^{n-1}(-1)^{n-i}\binom{n}{i}+\binom{n}{n}=0 \therefore (\color{red}{*})\sum\limits_{i=0}^{n-1}(-1)^{n-i}\binom{n}{i}=-1$$ because $$\binom{n}{n}=1$$.

Let

$$f(x)=(x-1)^n-x^n=\binom{n}{0}(-1)^n+\binom{n}{1}(-1)^{n-1}x+\cdots +\binom{n}{n-1}(-1)^1x^{n-1}$$,

$$\frac{d}{dx}[f(x)]=\frac{d}{dx}[(x-1)^n-x^n]=n(x-1)^{n-1}-nx^{n-1}=n((x-1)^{n-1}-x^{n-1})\leq 0$$ for $$x\geq1$$.

Note that $$f(1)=-1$$ by above $$(\color{red}{*})$$, and $$f$$ is decreasing, thus $$f(x)\leq -1$$ for $$x\geq 1$$.

$$\therefore (x-1)^n=f(x)+x^n\leq -1+x^n$$ for $$x\geq 1, n\in \mathbb{N}$$.

Here's a proof without induction that allows $$n$$ to be any real number $$\gt 1.$$ Let $$f(x)= x^n -1$$ and $$g(x)= (x-1)^n.$$ Observe that $$f'(x)= nx^{n-1}$$ and $$g'(x)= n(x-1)^{n-1}.$$ Since $$f(1)=g(1)$$ and $$f'(x) \gt g'(x) \text{ whenever } x \gt 1,$$ it follows that $$f(x) \gt g(x) \text{ whenever } x \gt 1.$$