= Solution
The <Fermat-Euler theorem> states that, for a positive <integer> $q$ and an <integer> $a$ coprime to $q$, $a^{\varphi(q)}\equiv1\pmod q$, where the <Euler totient function> $\varphi(q)$ counts residue classes coprime to $q$. The case $q=1$ is trivial. For $q>1$, let $r_1,\ldots,r_{\varphi(q)}$ be a <reduced residue system>. Multiplication by $a$ permutes it: equality of two resulting residues can be cancelled using the modular inverse of $a$, and the products remain coprime to $q$. Multiplying the residues before and after the permutation gives
$$
a^{\varphi(q)}r_1\cdots r_{\varphi(q)}\equiv r_1\cdots r_{\varphi(q)}\pmod q.
$$
Every factor on the right is a unit, so its product can be cancelled. This proves
$$
\boxed{a^{\varphi(q)}\equiv1\pmod q.}
$$
The <integer> $359$ is <prime>: none of the <primes> $2,3,5,7,11,13,17$ up to $\sqrt{359}<19$ divides it. Therefore $\varphi(359)=358$. Apply the theorem to $60$ and use the supplied congruence:
$$
10^{179}\equiv(60^2)^{179}=60^{358}\equiv1\pmod{359}.
$$
To connect this to individual digits rather than merely assert periodicity, use long-division remainders $r_0=1$ and
$$
10r_{n-1}=359a_n+r_n,\qquad0\le r_n<359.
$$
Induction gives $r_n\equiv10^n\pmod{359}$. Since $10$ is coprime to $359$, no remainder is zero. The congruence above therefore implies $r_{n+179}=r_n$ for every $n\ge0$, using the unique representative in $1,\ldots,358$. In particular
$$
a_{n+179}=\left\lfloor\frac{10r_{n+178}}{359}\right\rfloor
=\left\lfloor\frac{10r_{n-1}}{359}\right\rfloor=a_n.
$$
Hence
$$
\boxed{a_{n+179}=a_n\quad(n\ge1).}
$$
This is the mechanism of <decimal period from multiplicative order>: the period divides $179$.
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