Irreducibility of the standard representation of $S_n$. [duplicate]

The permutation representation of $S_n$ is $\mathbb C^n$ with elements of $S_n$ permuting the basis vectors $\{e_1, e_2, \ldots, e_n\}$. It has a trivial subrepresentation spanned by the vector $v = \sum_i e_i$. By Maschke's theorem there is a complement subrepresentation (given by the condition $\sum_i x_i = 0$, where $x_i$ is the i'th coordinate). This last representation (the standard representation of $S_n$) is irreducible. Why is this?

If we try characters, we can get the following: denote the standard representation by $V$ (dimension n-1), and the full permutation representation by $Perm$. Then $$ Perm = V\oplus Triv, $$ where the latter is the trivial representation. This means that $\chi_V(g) = |X^g| - 1$, and $$ \langle\chi_V, \chi_V\rangle = \frac{1}{|G|}\sum(|X^g|-1)^2|C(g)| = 1+\frac{1}{|G|}\sum (|X^g|^2 - 2|X^g|)|C(g)|, $$ where $X = \{1,2,\ldots, n\}$ and $X^g$ is the set of fixed points of permutation $g\in S_n$ and $C(g)$ is the conjugacy class of $g$. The sum is taken over all different conjugacy classes of $S_n$.

Now, since I know that $V$ is irreducible, the last sum must be zero. I don't see, however, why that should be.

Perhaps there is a direct proof or irreducibility?

Thank you.

The question seems to be answered in comments, so I am posting this CW-answer, so that it does not remain unanswered.

Qiaochu posted link to this MO thread where a proof using characters can be found. He also mentioned that there is a simple solution not using characters, which is a nice exercise; so you might want to try to do the proof yourself. I have tried to write down this simple solution in an answer to another question.