For the Fibonacci numbers, show for all $n$: $F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}$ The definition of a Fibonacci number is as follows:
$$F_0=0\\F_1=1\\F_n=F_{n-1}+F_{n-2}\text{ for } n\geq 2$$ 
Prove the given property of the Fibonacci numbers for all n greater than or equal to 1.
$$F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}$$
I am pretty sure I should use weak induction to solve this. My professor got me used to solving it in the following format, which I would like to use because it help me map everything out...
This is what I have so far:
Base Case: Solve for $F_0$ and $F_1$ for the following function: $F_nF_{n+1}$.
Inductive Hypothesis: 
What I need to show: I need to show $F_{n+1}F_{n+1+1}$ will satisfy the given property.
Proof Proper:  (didn't get to it yet)
Any intro. tips and pointers?

This identity is clear from the following diagram:

(imagine here a generalized picture with $F_i$ notation)
The area of the rectangle is obviously
$$F_n(F_{n}+F_{n-1})=F_nF_{n+1}$$
On the other hand, since the area of a square is x^2, it is obviously:
$$F_1^2+F_2^2+\dots+F_n^2$$
Therefore:
$$F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}$$

You can even convert this graphical proof to an inductive proof - your inductive step would consist of adding a square $F_{n+1} * F_{n+1}$.
