Brian Bi
$\DeclareMathOperator{\End}{End} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\span}{span} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\ad}{ad} \newcommand\d{\mathrm{d}} \newcommand\pref[1]{(\ref{#1})}$

Remark: I did not understand the proof of Lemma 5.13.3 given in the text, so I consulted Fulton & Harris, which cited Curtis & Reiner. It appears that the key is to consider the linear operator $$T$$ on $$\mathbb{C}[S_n]$$ that acts by right-multiplication by $$c_\lambda$$. This was not clear in Etingof's proof.

Explicitly, let $$\gamma$$ denote the scalar such that $$c_\lambda^2 = \gamma c_\lambda$$. Since the coefficient of the identity element in $$c_\lambda$$ is 1, computing the trace of $$T$$ in the basis consisting of the elements of $$S_n$$ yields $$n!$$. We now compute $$\tr T$$ in a second way. Write $$\mathbb{C}[S_n] = V_\lambda \oplus W$$, a direct sum of vector spaces. (Note: the argument used in the proof of Maschke's theorem implies that it's possible to choose $$W$$ to be a subrepresentation of $$\mathbb{C}[S_n]$$, but we won't need that fact here.) Now $$n! = \tr T = \tr(T|_{V_\lambda}) + \tr(T|_W)$$. For $$x \in V_\lambda$$, we have $$x = yc_\lambda$$ for some $$y \in \mathbb{C}[S_n]$$, so that $$T(x) = xc_\lambda = yc_\lambda^2 = \gamma yc_\lambda = \gamma x$$. Thus, $$T|_{V_\lambda} = \gamma \Id$$ and the trace on this block is $$\gamma \dim V_\lambda$$. On the other hand, for arbitrary $$x \in \mathbb{C}[S_n]$$, we have $$T(x) \in V_\lambda$$, so $$\tr(T|_W) = 0$$. This implies that $$n! = \gamma \dim V_\lambda$$, or $$\gamma = \frac{n!}{\dim V_\lambda}$$ as required.