Proving that $x^{16} > 5$ when given a polynomial of degree $15$. I am unable to prove the following

If $x^{15}-x^{13}+x^{11}-x^9+x^7-x^5+x^3-x = 7$ prove that $x^{16} > 15$.


Notice that the left hand is equal to
$(x^3-x)(x^{16}-1)/(x^4-1)$; hence the equation will be transfer into:
$(x^{16}-1)=7(x^4-1)/(x^3-x)=7(x^2+1)/x$; notice that $(x^2+1)/x$ is graeter than 2, hence the right hand is greater than 7*2, therefor $x^{16}>15$.
