Intuitive proof that $U(n)$ isn't isomorphic to $SU(n) \times S^1$ One way to prove this is by comparing their centers. However, I do not feel that this proof gives me much insight into the structures of the groups. (It would make me very happy if I were to be corrected though!)
Is there a more intuitive proof? My intuition is that the second group is "larger", but I haven't been able to turn this into a proof.
 A: $n \ge 2$ throughout. Taking determinants gives a short exact sequence
$$1 \to SU(n) \to U(n) \xrightarrow{\det} U(1) \to 1$$
and what you'd really like is for an isomorphism $U(n) \cong SU(n) \times U(1)$ to respect the structure of this short exact sequence. (If there were some random isomorphism that didn't have this property that would be less interesting.) 
For starters, this requires that $det : U(n) \to U(1)$ have a section, or equivalently that the short exact sequence above splits, which it does: for example, we can send $z \in U(1)$ to the diagonal matrix with entries $z, 1, 1, \dots$. This section allows us to write $U(n)$ as a semidirect product
$$U(n) \cong SU(n) \rtimes U(1)$$
and this cannot be improved to a direct product. This is possible if and only if it is possible to choose a section of $\det : U(n) \to U(1)$ whose image is central. Now, the center of $U(n)$ is the diagonal copy of $U(1)$. When restricted to the center, the determinant map is the map
$$U(1) \ni z \mapsto z^n \in U(1)$$
and this map doesn't have a section for the same reason that $\sqrt[n]{z}$ is not a holomorphic function of $z$. 
