= Solution
Let $|\psi\rangle=a|0\rangle+b|1\rangle$ and $|\phi\rangle=c|0\rangle+d|1\rangle$ be normalized <qubit states>. By <linearity>, the <SWAP gate> gives
$$
\begin{aligned}
\mathrm{SWAP}(|\psi\rangle\otimes|\phi\rangle)
&=ac|00\rangle+ad|10\rangle+bc|01\rangle+bd|11\rangle\\
&=(c|0\rangle+d|1\rangle)\otimes(a|0\rangle+b|1\rangle)\\
&=\boxed{|\phi\rangle\otimes|\psi\rangle.}
\end{aligned}
$$
The output remains a <product state>. Single-<qubit> <unitary operators> also preserve <product states>, so an induction over any <quantum circuit> made from these operations shows that it cannot create <quantum entanglement> from a <product state>. Equivalently, $\mathrm{SWAP}(A\otimes B)=(B\otimes A)\mathrm{SWAP}$ moves every exchange through the local operations, leaving only local gates and a permutation of wires. This is the <product-state obstruction to SWAP universality>.
Consequently \b[<SWAP gates> and arbitrary single-<qubit> gates do not form a <universal quantum gate set>]. For example, they cannot map $|00\rangle$ to the <Bell state> $(|00\rangle+|11\rangle)/\sqrt2$, whereas a <Hadamard gate> followed by a <CNOT gate> does so.
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