A question on the proof of $\text{pf}(MAM^T) = \text{det}(M) \cdot \text{pf}(A)$ There is a step in the proof of pf($MAM^T$) = det($M$) pf($A$) for a skew symmetric matrix $A$ that I do not understand. The proof I can find basically says that $$\text{det}(MAM^T) = (\text{det} M)^2 \text{det}(A),$$ thus $$\text{pf}(MAM^T)^2 = (\text{det} M)^2 (\text{pf} A)^2.$$ and by taking square root of both sides we obtain $$\text{pf}(MAM^T) = \pm (\text{det} M) \text{pf} A.$$ Then the proof ends by substituting $M=I$ and concludes the sign must be positive. My question is, isn't it possible that the sign can be sometimes positive and sometimes negative, hence substituting $M=I$ is not suffice to show that the sign is always positive?

Hint: if $\det A\
e 0$, $$f(M) := \frac {\operatorname{pf}(MAM^T)} {\operatorname{pf}A\cdot \det M}$$ is well-defined and continuous on $GL(n,\mathbb C)$.