$S_n$ acts on $\mathbb{C}^n$ by permutation, and there are two conventions which work (permute basis vectors or permute components), but permuting basis vectors ends up being a bit more natural. Here is a little background first. Given a group $G$ and two sets $A$ and $B$ which $G$ acts on, then the set of functions $A\to B$ is acted upon by $G$ using the action defined by $(gf)(a)=gf(g^{-1}a)$. The inverse makes it so $g(hf)=(gh)f$.
A vector is a function $v:[n]\to \mathbb{C}$, and viewing $[n]=\{1,2,\dots,n\}$ as a set with an $S_n$ action (in particular, the defining action of $S_n$) and viewing $\mathbb{C}$ as the trivial representation, $S_n$ acts on vectors by $(\sigma v)_i=v_{\sigma^{-1}(i)}$. On standard basis vectors, one can verify that $\sigma e_i=e_{\sigma(i)}$:
$$(\sigma e_i)_j=\delta_{i,\sigma^{-1}(j)}=\delta_{\sigma(i),j}=(e_{\sigma(i)})_j$$
Let $V$ be the span of the vectors you describe, so $V$ is the orthogonal complement of the vector $(1,1,\dots,1)$, or in other words the set of vectors whose components sum to $0$. Certainly, permuting the entries of a vector will not change whether the components sum to $0$, so $V$ is a subrepresentation of dimension $n-1$.
That $V$ is simple can be shown by showing that it is cyclic for any nonzero $v\in V$. Just give a process which can take an arbitrary vector and give one of the basis vectors through a sequence of permuting entries and linear combinations, then show that every basis vector can be reached from that basis vector.
Another way is to calculate the character $\chi$ of the representation and show that $(\chi,\chi)=1$. (Easiest by starting with the character of the $\mathbb{C}^n$ representation and subtracting off the trivial representation.)
