Shapley value: an alternative representation

It is my belief that the more common representation of the Shapley value is given by $$\phi_i(v)=\sum_{S\subseteq N-i} \frac{|S|!(|N|-|S|-1)!}{|N|!}(v(S\cup\{i\})-v(S))$$ where $v \in \mathbb{R}^{L(N)}$ is a coalitional game on the finite player set $N$, and $i\in N$. (Note $L(N)=\{S|S\subseteq N \text{ and } S\ne\emptyset\}$.)

There is yet another representation of the Shapley value, claimed to be equivalent to the previous, given by $$\phi_i(v)=\sum_{S\subseteq N-i} \frac{|S|!(|N|-|S|-1)!}{|N|!}(v(N\backslash S)-v(S)).$$ (See R. B. Myerson, Game Theory, 1991, p.441, and also an online note here on p. 10.)

Frankly, I do not see how these two formulae are equal. A naive yet natural sufficient condition for these to be equivalent is for $v(S\cup\{i\})=v(N\backslash S)$, which is not generally true. May anyone enlighten me? Thank you!

The piece with $v(S \cup \lbrace i \rbrace)$ in the first formula, is equal to the same thing with $v(N \backslash T)$ where $T$ is the complement of $S$ in $N - \lbrace i \rbrace$. The sum can be indexed by $T$ instead of $S$, and will give the same result, because the weight of $S$ and $T$ (the expression with factorials) is the same by the equation ${{N-1} \choose |S|} = {{N-1} \choose {N - 1 - |S|}}$.
The second term, with "$- v(S)))$", is the same in both formulas.