= Solution
The equality $\sum_\lambda(\dim S^\lambda)^2=n!$ exhausts the <dimension> of the <regular representation>. There is no room for another irreducible constituent, since it would contribute its positive squared <dimension>. Thus
$$
\boxed{\{S^\lambda:\lambda\vdash n\}\text{ is a complete irredundant family of irreducible representations in characteristic zero}.}
$$
Each has the standard-<polytabloid> <basis> just proved. The usual split characteristic-zero statement follows over any characteristic-zero <field> because these integral Specht models remain irreducible after extension to an algebraic closure and already exhaust all dimensions there. In the literal wording of the source, <polytabloids> themselves are vectors; it is the <Specht modules> with those bases that form the family of <representations>.
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