The formula does not come from magic. Here's how I think about it. Suppose you give the Young diagram corresponding to the partition $d = (d-1) + 1$ the standard tableaux of $1,2,3,\ldots, d-1$ on the first row and $d$ for the second row. Now instead of applying the young symmetrizer straight up to $\Bbb{C}[S_d]$ you can first apply $a_\lambda$ and see what happens. Now we claim that $\Bbb{C}[S_d]a_\lambda$ is $d$ dimensional. To see this, first notice that that the row group $P_\lambda \cong S_{d-1}$.
Now for any $e_g,e_h$ with $g,h\in S_{d-1}$, we have $e_g a_\lambda = e_ha_\lambda$. This is because $a_\lambda$ is the sum of all elements in $P_\lambda\cong S_{d-1}$ and multiplying again by an $e_g$ for $g \in S_{d-1}$ just permutes the order of summation in $a_\lambda$. More generally, we see that for any $e_g, e_h \in \Bbb{C}[S_d]$ such that $g^{-1}h \in S_{d-1}$, we have
$$e_ga_\lambda = e_ha_\lambda.$$
This comes down to the fact that two left cosets $gS_{d-1}$ and $hS_{d-1}$ are equal iff $g^{-1}h \in S_{d-1}$. Hence $\Bbb{C}[S_d]a_\lambda$ has basis vectors $v_i$ that are
$$v_i = e_\sigma a_\lambda$$
where $\sigma$ is a 2-cycle of form $(i\hspace{1mm} d)$ for $1 \leq i \leq d$ with the convention that $(d\hspace{1mm} d)$ is the identity.
The final step in the problem is to apply $b_\lambda$ to each of these basis vectors and show that their total sum is zero. Indeed, this can be seen as follows. We write $\big(\sum_{i=1}^d v_i\big)b_\lambda=\big(\sum_{i=1}^d v_i\big)\big(1-(1\ d)\big)$ as
$$\begin{array}{ccccccc}\bigg(e_{(1)} a_\lambda &+& e_{(1d)}a_\lambda &+& e_{(2d)}a_\lambda &+& \ldots &+& e_{(d-1 \hspace{1mm} d)} a_\lambda \bigg) \\
&&&& \text{minus} &&\\
\bigg(e_{(1)} a_\lambda e_{(1d)} &+& e_{(1d)}a_\lambda e_{(1d)} &+& e_{(2d)}a_\lambda e_{(1d)} &+& \ldots &+& e_{(d-1 \hspace{1mm} d)} a_\lambda e_{(1d)}\bigg).\end{array}$$
Notice we can decompose $e_{(1)}a_\lambda$ as
$$e_{(1d)} \left(\sum_{ g\in P_\lambda, g(1) = 1} e_g \right) e_{(1d)} + e_{(1d)} \left(\sum_{ g\in P_\lambda, g(2) = 1 } e_g \right) e_{(2d)} + \ldots + e_{(1d)} \left(\sum_{ g\in P_\lambda, g(d-1) = 1 } e_g \right) e_{(d-1\hspace{1mm} d)}. $$
A similar decomposition exists for elements in the first row in that big fat expression we wrote for $\sum_{i=1}^d v_ib_\lambda$. You should be able to see now that the sum is in fact zero, so that $\Bbb{C}[S_d]c_\lambda$ is spanned by
$$v_2b_\lambda,\hspace{1mm} v_3b_\lambda, \hspace{1mm} \ldots, v_db_\lambda.$$
However each of these vectors is precisely the $v_j$ that they have in the answer at the back of Fulton and Harris so we are done.
Instead of $\mathbf{i}$ I'll use $v_i$. This just amounts to showing that $$\operatorname{Span}_{\mathbb C}(v_i-v_j:1\leq i<j\leq n)=\left\{\sum_{i=1}^n c_iv_i:\sum_{i=1}^n c_i=0\right\}.$$
The forward inclusion is clear since $v_i-v_j$ has coefficients of the $v_k$'s which sum to zero for each $1\leq i<j\leq n$. Now try to write an arbitrary element $\sum_{i=1}^n c_iv_i$ with $\sum_{i=1}^n c_i=0$ as a sum of $v_i-v_j$ terms. For example, you can use $c_n=-c_1-\cdots-c_{n-1}$ so that $$\sum_{i=1}^n c_iv_i=c_1(v_1-v_n)+\cdots+c_{n-1}(v_{n-1}-v_n).$$
Best Answer
Let $\Phi:\mathbb{C}[S_n]\rightarrow \mathbb{C}[S_n]$ be the linear map which sends a permutation $\sigma$ to ${\rm sgn}(\sigma)\ \sigma$. It is an algebra automorphism of $\mathbb{C}[S_n]$ which maps the Specht module for $\lambda$ to that for the transpose $\lambda'$. Hint: Is this map $S_n$-equivariant? If not, how can you correct for that?