Proof by Induction: $(1+x)^n \le 1+(2^n-1)x$

Question

I have to prove the following by induction:

$$(1+x)^n \le 1+(2^n-1)x$$

for $n \ge 1$ and $0 \le x \le 1$.

I start by showing that it's true for $n=1$ and assume it is true for one $n$.

$$(1+x)^{n+1} = (1+x)^n(1+x)$$ by assumption: $$\le (1+(2^n-1)x)(1+x)$$ $$= 1+(2^n-1)x+x+(2^n-1)x^2$$

That is what I was able to do on my own. I do not understand the next step of the solution (they are now using $x$ instead of $x^2$ at the end):

$$Because\,0 \le x \le 1$$ $$(1+x)^{n+1} \le 1 + (2^n-1)x+x+(2^n-1)x$$ $$=1+2(2^n-1)x+x$$ $$=1+(2^{n+1}-1)x$$

Because $0 \le x \le 1$ I assume that $(2^n-1)x^2 < (2^n-1)x$. So in my opinion, I just proved that $(1+x)^{n+1}$ is less than something greater than what I had to prove initially. Why is it enough to prove the original problem?

Answer 1

I'm not exactly clear as to what your final sentence means. But, since $0\le x\le 1$, we know that $x^2\le x\cdot x\le 1\cdot x\le x$, so that $$(1+x)^{n+1} \le 1 + (2^n-1)x + x + (2^n-1)x^2 \le 1 + (2^n-1)x + x + (2^n-1)x = 1+2(2^n-1)x + x = 1 + (2^{n+1}-1)x,$$ which is the statement for $n+1$.

Answer 2

$$(1+x)^{n+1}=(1+x)(1+x)^{n}\le(1+x)(1+(2^{n}-1)x)=1+(2^{n}-1)x+x(1+(2^{n}-1)x)$$

$$\underbrace{\le}_{x\le1}1+(2^{n}-1)x+x(1+2^{n}-1)=1+(2^{n}-1)x+2^{n}x$$

$$=1+(2\cdot2^{n}-1)x=1+(2^{n+1}-1)x$$