Proving that $g(1) = 1$, where $g$ is a multiplicative arithmetic function

Question

I'm having some trouble understanding a simple problem about an arithmetic function. The problem is simply to answer true or false that

$g(1) = 1$, assuming $g$ is multiplicative and $g(n)$ $\neq 0$ for any $n$.

My work:

Since $g$ is multiplicative,

$$g(a\cdot1) = g(a)g(1)$$

$$g(a) \quad\,= g(a)g(1)$$

$$1 \;\,\, = g(1)$$

So $g(1)=1$ is a true statement.

I suspect I have oversimplified the problem, but I cannot yet understand how.

Answer 1

$g(1)\cdot g(1)=g(1\cdot 1)=g(1)$. Since $g(1)\not =0$ we can say $g(1)=1$.

Answer 2

In fact as long as there is at least one value $n_0$ for which $g(n_0)\ne 0$, we have $g(1)=1$.

(By the same argument: $g(n_0)=g(n_0\cdot 1)=g(n_0)\cdot g(1)$.
So $g(1)=1$.