$F(2n-1) = F(n-1)^2 + F(n)^2$, where $F(i) $ is the $i$'th Fibonacci number, for all natural numbers greater than $1$ I'm proceeding by induction, but I have no idea how to do the inductive step. The Fibonacci sequence takes on the definition:
$F(0)=0$
$F(1) = 1$
$F(2) = 1$
So far, my proof is as follows:
Base Case:
$F(2) = 1 = F(1)^2 + F(0)^2$. This case holds
Inductive Step: We assume the claim holds for $n$ and show it holds for $n+1$...
$F(2n+1) = F(2n)+F(2n-1)=F(2n) + F(n-1)^2+F(n)^2$
I am stuck on how to reduce this expression to $F(n)^2 + F(n+1)^2$...
Edit: Thank you for the responses. I realize I made an egregious error in defining the fibonacci sequence. I have corrected it to the definition we use in my class.
Edit by DS : Now changed the indexing to the standard convention of the Fibonacci numbers.
 A: We need a second part to our inductive hypothesis
\begin{eqnarray*}
F_{2n-1}=F_{n-1}^2+ F_n^2 \\
F_{2n}=F_{n}(F_{n+1}+F_{n-1})
\end{eqnarray*}
To prove the first equation
\begin{eqnarray*}
F_{2n+1}=F_{2n}+F_{2n-1}= F_{n}(F_{n+1}+F_{n-1}) + F_{n-1}^2+ F_n^2 \\= F_{n-1}(F_{n-1}+F_n)+F_n^2+F_nF_{n+1}= F_{n+1}^2+F_{n}^2.
\end{eqnarray*}
To prove the second equation
\begin{eqnarray*}
F_{2n+2}=F_{2n+1}+F_{2n}=  F_{n+1}^2+F_{n}^2+F_{n}(F_{n+1}+F_{n-1}) \\ =F_{n+1}(F_{n+1}+F_{n})+F_{n}(F_{n}+F_{n-1})= F_{n+1}(F_{n+2}+F_{n}).
\end{eqnarray*}
A: Yet another approach. By setting $M=\small\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}$ we have
$$ \begin{pmatrix} F_{n+2} \\ F_{n+1}\end{pmatrix} = M \begin{pmatrix} F_{n+1} \\ F_{n}\end{pmatrix},\qquad M^n=\begin{pmatrix}F_{n+1} & F_n\\ F_{n} & F_{n-1}\end{pmatrix} $$
and $F_{2n+1}$ is the top-left element of $M^{2n}=(M^n)^2$, where
$$ \begin{pmatrix}F_{n+1} & F_n\\ F_{n} & F_{n-1}\end{pmatrix}^2 = \begin{pmatrix}\color{blue}{F_{n+1}^2+F_n^2} & F_n(F_{n-1}+F_{n+1})\\ F_n(F_{n-1}+F_{n+1}) & F_n^2+F_{n-1}^2\end{pmatrix} $$
leads to
$$ F_{2n+1}=F_{n+1}^2+F_n^2,\qquad \boxed{F_{2n\color{red}{-1}}=F_n^2+F_{n-1}^2.}$$

By Binet's explicit formula we have $F_{2n}=F_n L_n$, where Lucas numbers $L_n$ are defined through $L_0=2,L_1=1$ and $L_{n+2}=L_{n+1}+L_n$.
A: A different approach:
The key idea is to prove a more general statement. With the initial statement, we can see that odd Fibonacci numbers seem to be quite annoying to work with.
The hypothesis is
$$ F_n = F_{n-k}F_k + F_{n-k-1}F_{k-1} $$
for well-chosen $k$ (You can work out the details.). Now this identity is easy to prove via induction and we can plug in $n = 2k$ to get the result.
