Fibonacci's final digits cycle every 60 numbers How would you go about to prove that the final digits of the Fibonacci numbers recur after a cycle of 60? 
References:
The sequence of final digits in Fibonacci numbers repeats in cycles of 60. The last two digits repeat in 300, the last three in 1500, the last four in , etc. The number of Fibonacci numbers between  and  is either 1 or 2 (Wells 1986, p. 65).
http://mathworld.wolfram.com/FibonacciNumber.html

Note that
$$
\begin{pmatrix} F_{n+1}\\ F_{n+2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix}
$$
and
$$
\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^{60} \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \mod 10.
$$
One can verify that $60$ is the smallest power for which this holds, so it is the order of the matrix mod 10.  In particular
$$
\begin{pmatrix} F_{n+60}\\ F_{n+61} \end{pmatrix} \equiv \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix} \mod 10
$$
so the final digits repeat from that point onwards.
