Francois Ziegler's answer is not massive overkill. The proof is simple. 

Suppose you have a continuous multiplicative mapping 
$P: \operatorname\{Mat\}_n(\mathbb R) \to (\mathbb R, \cdot)$ as you started with, then it restricts to a continuous group homomorphism 
$P:GL(n)\to (\mathbb R\setminus\{0\}, \cdot)$, which is analytic (using $\exp$).
Its derivative at $\mathbb I_n$ is a Lie algebra homomorphism
$P':\mathfrak g\mathfrak l(n)\to \mathbb R$ which must vanish on each commutator.
The space of all commutators is the codimension 1 Lie subalgebra $\mathfrak s\mathfrak l(n)$.
Since $P'$ is also linear, it is of the form $P'(X) = k.\operatorname\{Trace\}(X)$ for some $k$. 
This integrates to $P(A) = \det(A)^k$. Here $k$ must be integral if the ground field is $\mathbb C$. In the real case any $k$ works if $\det(A)$ is always $\ge 0$, and integral generally.