Symmetric and exterior powers of a projective (flat) module are projective (flat)

Assume that $R$ is a commutative ring with unity and $P$ a projective (flat) $R$-module. Why $\mathrm{Sym}^n(P)$ and $\Lambda^n(P)$ are projective (flat) for every $n$?

As Darij remarked in his comment, if $R$ is not a ${\mathbb Q}$-algebra one cannot argue that $\bigwedge^n P$ and ${\mathfrak S}^n P$ are summands of $P^{\otimes n}$, thereby reducing the statement to showing that projectivity and flatness are stable under taking tensor powers. Instead, one might argue as follows:

1. Freeness: If $F$ is a free $R$-module on with basis $\{v_1,\ldots,v_n\}$, then $\bigwedge^k F$ is free over $\{v_{i_1}\wedge\cdots\wedge v_{i_k}\ |\ i_1<\ldots<i_k\}$ and ${\mathfrak S}^n F$ is free over $\{v_{i_1}\cdots v_{i_k}\ |\ i_1\leq \ldots\leq i_k\}$. The same holds for a possibly infinitely generated free module $F\cong R^{(I)}$ if $I$ is equipped with a linear order.

   Side question: What if we do not impose $\textsf{AC}$? Are $\bigwedge^n R^{(I)}$ and ${\mathfrak S}^n R^{(I)}$ still free then?

2. Projectivity: The projective modules are the summands/retracts of free modules. Since $\bigwedge^n(-)$ and ${\mathfrak S}^n(-)$ preserve retracts (as does any functor) and freeness, they therefore also preserve projectivity.

3. Flatness: The flat modules are the direct limits of finite rank free modules. Since and $\bigwedge^n(-)$ and ${\mathfrak S}^n(-)$ preserve direct limits and finite rank free modules, they therefore also preserve flatness.

Here's one approach using the fact that projective modules are direct summands of free modules. Suppose $P\oplus Q\cong R^k$. Then $$R^{k^n}=(P\oplus Q)^{\otimes n}=P^{\otimes n}\oplus\cdots ,$$ so $P^{\otimes n}$ is projective. But then notice that $\Lambda^n(P)$ and $\mathrm{Sym}^n(P)$ are both direct summands of $P^{\otimes n}$.

Edit: As Darij Grinburg points out, and I only just now noticed, this only works for some rings $R$, such as $\mathbb Q$ algebras.