= Solution
The <Fermat-Euler theorem> states that for integers $m\ge1$ and $a$ with $\gcd(a,m)=1$,
$$
\boxed{a^{\varphi(m)}\equiv1\pmod m,}
$$
where <Euler's totient function> $\varphi(m)$ counts the residue classes coprime to $m$. The case $m=1$ is trivial. For $m\ge2$, list a reduced residue system $r_1,\ldots,r_{\varphi(m)}$. Multiplication by $a$ permutes these classes: each $ar_i$ is coprime to $m$, and $ar_i\equiv ar_j$ implies $r_i\equiv r_j$ since $a$ has a multiplicative inverse modulo $m$. Multiplying the permuted residues gives
$$
a^{\varphi(m)}\prod_i r_i\equiv\prod_i r_i\pmod m.
$$
The product is itself coprime to $m$ and can be cancelled using its inverse, proving the theorem.
The integer 359 is prime: none of the primes $2,3,5,7,11,13,17$, all the primes not exceeding $\sqrt{359}<19$, divides it. Therefore $\varphi(359)=358$ and 60 is coprime to 359. Using the supplied congruence and the <Fermat-Euler theorem>,
$$
\boxed{10^{179}\equiv(60^2)^{179}=60^{358}\equiv1\pmod{359}.}
$$
Now express the decimal expansion by long division. Put $r_0=1$ and for $n\ge1$ define
$$
10r_{n-1}=359a_n+r_n,\qquad 0\le a_n\le9,\quad0\le r_n<359.
$$
Thus $r_n\equiv10^n\pmod{359}$, and none of these remainders is zero. The congruence $10^{179}\equiv1$ implies $r_{n+179}\equiv r_n$; since both are the unique representatives between zero and 358, they are equal. A digit is determined by the preceding remainder, namely $a_n=\lfloor10r_{n-1}/359\rfloor$, so
$$
\boxed{a_{n+179}=a_n\quad\text{for every }n\ge1.}
$$
This proves the required pure periodicity from the first digit. In fact its least period is 179: the <multiplicative order> of 10 modulo 359 divides the prime 179 and is not one, since $10\not\equiv1\pmod{359}$. The connection between the remainder recurrence and the period is the <decimal period from multiplicative order>.
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