Is every $N$th Fibonacci number where $N$ is divisible by $5$ itself divisible by $5$ There seems to be a pattern of sorts in the Fibonacci sequence:
The $5$th, $10$th, $15$th & $20$th values are:
$$5, 55, 610, 6765$$
Does this pattern continue ad infinitum? I've tried a few more using Binet's Formula and it seems to hold. So:

Is every $N$th Fibonacci number where $N$ is divisible by $5$ itself divisible by $5$?
Bonus Q: Are there any other patterns?


Yes, if $5$ divides $n$ then $5$ divides $F_n$. Actually also the other way round is true, so $5$ divides $n$ if and only if $5$ divides $F_n$.
More generally, the following congruence holds 
$$F_n\equiv n3^{n-1}\pmod{5}.$$ 
This congruence implies our claim and more than that. For example, it follows that the Fibonacci sequence modulo $5$ is periodic with period length $5\cdot 4=20$ where $5$ is the period of $n$ and $4$ the period of $3^{n-1}$  (recall that by Fermat's little theorem $3^{k(5-1)}\equiv 1 \pmod{5}$). The list of values $F_n\pmod{5}$ is
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
n \pmod{20}& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\ 
F_n\pmod{5} &\mathbf{0}& 1& 1& 2& 3& \mathbf{0}& 3& 3& 1& 4\\
\hline
n \pmod{20}&10& 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18&19\\ 
F_n\pmod{5} &\mathbf{0}& 4& 4& 3& 2& \mathbf{0}& 2& 2& 4& 1
\\\hline
\end{array}$$ 
Proof of the above congruence $F_n\equiv n3^{n-1}\pmod{5}$. By the  Binet's formula and the binomial theorem we have that
$$\begin{align}F_n&=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)\\
&=\frac{1}{2^{n}\sqrt{5}}\sum_{k=0}^{n}\binom{n}{k}\left((\sqrt{5})^k-(-\sqrt{5})^k\right)\\
&=\frac{1}{2^{n}\sqrt{5}}\sum_{j=0}^{\lfloor (n-1)/2\rfloor}\binom{n}{2j+1}\left((\sqrt{5})^{2j+1}-(-\sqrt{5})^{2j+1}\right)\\
&=\frac{1}{2^{n}\sqrt{5}}\sum_{j=0}^{\lfloor (n-1)/2\rfloor}\binom{n}{2j+1}(2\sqrt{5}\cdot 5^j)\\
&=\frac{1}{2^{n-1}}\sum_{j=0}^{\lfloor (n-1)/2\rfloor}\binom{n}{2j+1}5^j
\equiv 3^{n-1}\sum_{j=0}^{0}\binom{n}{2j+1}5^j\equiv n3^{n-1}\pmod{5}.
\end{align}$$
