AoPS Wiki talk:Problem of the Day/September 7, 2011

Solution

Raising that thing to the seventh doesn't look appealing, nor does solving for $x$. So we try getting to $x^7+\frac{1}{x^7}$ gradually.

Since $x+\frac{1}{x}=1$, $\left(x^2+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}=1$ and thus $x^2+\frac{1}{x^2}=-1$.

We can also compute: \[1(-1)=\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)=x^3+\frac{1}{x^3}+x+\frac{1}{x}=x^3+\frac{1}{x^3}+1\] so $x^3+\frac{1}{x^3}=-2$.

Squaring $x^2+\frac{1}{x^2}$, we see \[1=\left(x^2+\frac{1}{x^2}\right)^2=x^4+\frac{1}{x^4}+2\] and thus $x^4+\frac{1}{x^4}=-1$. Thus, \[-1(-2)=\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)=x^7+\frac{1}{x^7}+x+\frac{1}{x}=x^7+\frac{1}{x^7}+1\] and therefore, $x^7+\frac{1}{x^7}=2-1=\boxed{1}$.