I am trying to prove that
$ \operatorname{fib}(n)<\left(\frac{5}{3}\right)^n $
where $\operatorname{fib}(n)$ is the $n^{th}$ fibonacci number. For a proof I used induction, as we know
$\operatorname{fib}(1)=1,\, \operatorname{fib}(2)=1,\, \operatorname{fib}(3)=2$
and so on. So for $n=1$; $\operatorname{fib}(1)<\frac{5}{3}$, and for general $n>1$ we will have
$\operatorname{fib}(n+1)<\left(\frac{5}{3}\right)^{n+1}$ First of all, $\left(\frac{5}{3}\right)^{n+1} = \left(\frac{5}{3}\right)^{n} \cdot \left(\frac{5}{3}\right)$
We have $\operatorname{fib}(n)<\left(\frac{5}{3}\right)^n$ and $\operatorname{fib}(n+1)=\operatorname{fib}(n)+\operatorname{fib}(n-1)$, so by induction we would have $\operatorname{fib}(n-1)<\left(\frac{5}{3}\right)^{n-1}$, and because $\frac{5}{3}>1$, we have $\left(\frac{5}{3}\right)^{n}>1$, so we get the correct result. Are there any flaws in my proof?