Let $S_i(x_1,x_2,\dots,x_n)$ denote the $i$th elementary symmetric polynomial in $n$ variables. Is $S_i(1,2,\dots,p-2) \equiv 1 \pmod{p}$ for all values of $i$ from $0$ to $(p-2)$ whenever $p$ is prime?
Since $$\sum_{0}^{p-2} S_i(1,2,\dots,p-2) = (p-1)!,$$ it's obvious that $$\sum_{0}^{p-2}S_i(1,2,\dots,p-2)\equiv -1 \pmod{p}$$ I don't know, however, if this sum can ever be different from $(p-1)$.
(There's no need for the superscripts on elementary symmetric polynomials $e_i$.)
As Vieta's formula states, $e_i(x_1,\cdots,x_n)$ will be the $i$th coefficient of $(T+x_1)\cdots(T+x_n)$.
According to Euler's theorem, $x^{p-1}\equiv 1$ mod $p$ for all $x$, so the polynomial $T^{p-1}-1$ has every integer residue $1,\cdots,p-1$ as a zero. Since the integers mod $p$ form a field, zeros correspond to linear factors in any polynomial's factorization, hence $T^{p-1}-1\equiv (T-1)\cdots(T-(p-1))$.
Therefore mod $p$:
$$\begin{array}{ll} (T+1)\cdots(T+(p-2)) & \equiv(T-(p-1))\cdots(T-2) \\ & \equiv(T^{p-1}-1)/(T-1) \\ & \equiv T^{p-2}+\cdots+T^2+T+1.\end{array}$$
Evidently all the coefficients are $1$ mod $p$.