If x^15-x^13+x^11-x^9+x7-x^5+x^3-x = 7 prove x^16 > 15

Given x^15-x^13+x^11-x^9+x^7-x^5+x^3-x=7 , we are asked to show that x^16>15 :
First, note that x^15-x^13+x^11-x^9+x^7-x^5+x^3-x=x(x-1)(x+1)(x^4+1)(x^8+1) so the polynomial has real roots at -1,0, and 1.
For x<-1 the polynomial is negative.
For -1For 0So we know that x>1 for the polynomial to achieve 7.
Since x>1 we have (x-1)^2>0
==> x^2-2x+1>0
==>x^2+1>2x
==> (x^2+1)/x>2
Now multiply both sides of the equation by x^2+1 :
(x^2+1)(x^15-x^13+x^11-x^9+x^7-x^5+x^3-x)=7(x^2+1)
Multiplying and factoring we get:
x(x^16-1)=7(x^2+1)
Then:
x^16-1=(7(x^2+1))/x   ; but (x^2+1)/x>2 so
x^16-1>7(2) and
x^16>15 as required.
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x^16-1=(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)
and
x^15-x^13+x^11-x^9+x^7-x^5+x^3-x=
x(x-1)(x+1)(x^4+1)(x^8+1)
so multiplying the degree 15 polynomial by x^2+1 gives x times the degree 16 polynomial.