Finding growth bounds on Fibonacci Sequence I've been working on this following problem:

Find a constant $c< 1$ such that $F_n \leq 2^{cn}$ for all $n \geq 0$. 

I honestly have no idea where to begin on this. I've done plenty of proofs the one function is Big-O another, but I'm not sure where to start looking for c in this case. Guidance appreciated. Thanks!
 A: Binet's formula seems like overkill, given the (relatively weak) desired conclusion.
A more naive strategy is to pretend that $F_{n} = 2^{cn}$ and see what the recursion relation $F_{n+2} = F_{n} + F_{n+1}$ suggests. Here,
$$
2^{c(n+2)} = 2^{cn} + 2^{c(n+1)}, \quad\text{or}\quad
2^{2c} = 1 + 2^{c}.
$$
The latter is a quadratic in $2^{c}$, and by the quadratic formula has roots
$$
2^{c} = \frac{1 \pm \sqrt{5}}{2}.
$$
It's therefore natural to attempt to show $F_{n} \leq \bigl[(1 + \sqrt{5})/2\bigr]^{n}$ for $n \geq 0$; this is a straightforward induction.
A: Let $0\lt x=2^c\le 2$.
If $F_{n-2}\le x^{n-2}$ and $F_{n-1}\le x^{n-1}$ you have $F_n=F_{n-1}+F_{n-2}\le x^{n-1}+x^{n-2}$
So you will definitely succeed if you get the initial conditions (first values) to work and if  $x^{n-1}+x^{n-2}\le x^n$ which gives you $x^2-x-1\ge 0$ (dividing through by $x^{n-2}\gt 0$). For example, $x=\frac 53$ works for the inequality.
There has been some debate elsewhere about the initial values of $F_n$ - whether $F_0=0$ or $F_0=1$ - $F_0=0$ is the best choice, because then $F_n\mid F_{kn}$ for all positive integers $n$ and $k$. Then $x=\frac 53$ works for the initial values too.
Then you need to take logs to get the appropriate value of $c$. Note that $\varphi = \frac {1+\sqrt 5}2$ is the least value of $x$ which would satisfy the inequality.
A: By Binet's formula you may prove that:
$$0<F_n =\frac{1}{\sqrt{5}} \left(\frac{1+\sqrt{5}}2\right)^n-\frac{1}{\sqrt{5}} \left(\frac{1-\sqrt{5}}2\right)^n<\left(\frac{1+\sqrt{5}}2\right)^n$$ then choose $c$ such that 
$$
2^c=\frac{1+\sqrt{5}}2,
$$ giving
$$
c=\frac{\ln \left(\frac{1+\sqrt{5}}2\right)}{\ln 2}=0.6942...<1.
$$
A: Hint:
$$\text{If }\phi = \frac{1 + \sqrt5}{2}, \ F(n) = \frac{\phi^n - (1-\phi)^n}{\sqrt 5}\text{ where } F(n)\text{ is the nth Fibonacci number}$$
