= Solution
Suppose first that $f=O(g)$. Choose $C>0$ and $N$ so that $|f(n)|\leq C|g(n)|$ for $n\geq N$. There are only finitely many natural numbers below $N$. Because $g(n)\ne0$ at each of them, all ratios $|f(n)|/|g(n)|$ in this finite prefix are finite. Let
$$
c=\max\left(\{C\}\cup\left\{\frac{|f(n)|}{|g(n)|}:n<N\right\}\right).
$$
Then $|f(n)|\leq c|g(n)|$ both in the prefix and in the tail, hence on the entire <domain of a function>. This proves the <finite-prefix extension of a big O bound>.
Conversely, a bound valid for every natural number is certainly valid eventually. The purported real constant cannot be negative because $g$ never vanishes; if it is zero, enlarge it to a positive constant. The definition of <Big O notation> then applies. The nonvanishing assumption is what permits the finite exceptions to be absorbed into one constant.
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