Fibonacci Sequence Exercise I need some help checking the following solution. 
The Fib sequence is defined by $a_1 = 1, a_2 = 1$ and for all $n\geq 2$, $a_{n+1} = a_n + a_{n-1}$. Thus, the sequence begins: 
1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...
Prove that for all $n\geq 1$, $a_n <\left(\frac{5}{3}\right)^n$. 
So far I have: 
by induction 
$a_{n+1} = \left(\frac{5}{3}\right)^n+1 = a_n+a_{n-1} < \frac{5}{3}\cdot\left(\frac{5}{3}\right)^n $ from the definition and factoring. 
$\frac{5}{3}\cdot\left(\frac{5}{3}\right)^n  > a_n+a_{n-1} $
$\frac{5}{3}\cdot\left(\frac{5}{3}\right)^n  > a_n\cdot \frac{5}{3}$ from above 
$a_n\cdot \frac{5}{3} > a_n + a_{n-1}$
This is where I get stuck, is this a complete solution? Or is there further computation needed? 

We need  Strong induction here
Let   $a_n<\left(\frac53\right)^n$  for integer $1\le n\le m$
$\implies \displaystyle a_{m+1}=a_m+a_{m-1}<\left(\frac53\right)^m+\left(\frac53\right)^{m-1}=\left(\frac53\right)^{m-1}\left(\frac53+1\right)$
which will be $<\left(\frac53\right)^{m+1}$ if $\displaystyle \left(\frac53+1\right)<\left(\frac53\right)^2\iff 8\cdot3^2<3\cdot5^2$
