Fibonacci's final digits cycle every 60 numbers

Since each term in the Fibonacci sequence is dependent on the previous two, each time a $0\pmod{m}$ appears in the sequence, what follows must be a multiple of the sequence starting at $F_0,F_1,\dots=0,1,\dots$ That is, a subsequence starting with $0,a,\dots$ is $a$ times the sequence starting with $0,1,\dots$

Consider the Fibonacci sequence $\text{mod }2$: $$ \color{red}{0,1,1,}\color{green}{0,1,1,\dots} $$ Thus, the Fibonacci sequence repeats $\text{mod }2$ with a period of $3$.

Consider the Fibonacci sequence $\text{mod }5$: $$ \color{red}{0,1,1,2,3,}\color{green}{0,3,3,\dots} $$ Thus, the Fibonacci sequence is multiplied by $3\pmod{5}$ each "period" of $5$. Since $3^4=1\pmod{5}$, the Fibonacci sequence repeats $\text{mod }5$ with a period of $20=4\cdot5$.

Thus, the Fibonacci sequence repeats $\text{mod }10$ with a period of $60=\operatorname{LCM}(3,20)$.

Notice that:

$F_{n+15} \equiv 7F_n \pmod{10}$ for $n\geq 1$.

Also the order of $7$ mod $10$ is $4$ so the repetition in the digits of the Fibonacci numbers begins after place $15\times 4 = 60$.

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.