Show that $\left(\frac{1+\sqrt{5}}{2}\right)^5 > 10$ I have to show that $\left(\frac{1+\sqrt{5}}{2}\right)^5 > 10$.
I've already proven that if $f_n$ is the nth Fibonnaci number, then,
$$f_{n+1} > \left(\frac{1+\sqrt{5}}{2}\right)^{n-1} $$
But I really don't get how to go from this statement to what I need to prove.
Any advices on how to take from here? Any help is welcome.
Thanks.

Binomially expand
\begin{eqnarray*}
\left( \frac{1+ \sqrt{5}}{2} \right)^5 &=& \frac{1+5 \sqrt{5}  +10 \times 5 +10 \times 5 \sqrt{5}+5 \times 25 +25 \sqrt{5}}{32} \\
&=&\frac{11+5 \sqrt{5}}{2}.
\end{eqnarray*}
Now $125>121 $ square root this and we have the stronger result that 
\begin{eqnarray*}
\left( \frac{1+ \sqrt{5}}{2} \right)^5 > 11. \\
\end{eqnarray*}
