Note that both of our polynomials are bijective functions on $\mathbb{F}_{p}$: $f_{1}(x)=x+1$ is the cycle $0 \rightarrow 1 \rightarrow 2 \rightarrow \cdots \rightarrow(p-1) \rightarrow 0$ and $f_{2}(x)=x^{p-2}+x^{p-3}+\cdots+x^{2}+2x+1$ is the transposition $0 \leftrightarrow 1$ (this follows from the formula $f_{2}(x)=\frac{x^{p-1}-1}{x-1}+x$ and Fermat's little theorem). So any composition formed from them is also a bijection, and reduction modulo $x^{p}-x$ does not change the evaluation in $\mathbb{F}_{p}$. Also note that the transposition and the cycle generate the symmetric group ( $f_{1}^{k} \circ f_{2} \circ f_{1}^{p-k}$ is the transposition $k \leftrightarrow(k+1)$, and transpositions of consecutive elements clearly generate $S_{p}$ ), so we get all $p!$ permutations of the elements of $\mathbb{F}_{p}$. The set $W$ only contains polynomials of degree at most $p-1$. This means that two distinct elements of $W$ cannot represent the same permutation. So $W$ must contain those polynomials of degree at most $p-1$ which permute the elements of $\mathbb{F}_{p}$. By minimality, $W$ has exactly these $p!$ elements.