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

$$(1+x)^{n+1}=(1+x)(1+x)^{n}\le(1+x)(1+(2^{n}-1)x)=1+(2^{n}-1)x+x(1+(2^{n}-1)x)$$
$$\underbrace{\le}_{x\le1}1+(2^{n}-1)x+x(1+2^{n}-1)=1+(2^{n}-1)x+2^{n}x$$
$$=1+(2\cdot2^{n}-1)x=1+(2^{n+1}-1)x$$
