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.
Well a first observation is that if $p \vert n$ then the $n-1$ dimensional standard representation itself is not irreducible, and this will propagate to its exterior powers.
It turns out this is the only obstruction and that if $p \not\vert n$ then indeed these are indeed irreducible. One can probably prove this directly in this case, but I'll just appeal to a general theorem:
Carter's criterion (formulated in full by James and Mathas, with the last piece proved by Fayers) gives a combinatorial condition for which Specht modules $S^\lambda$ remain irreducible in characteristic $p$. I won't state it in full, but it involves looking at the $p$-valuations of the hook lengths. In the case of hook shape partitions the condition degenerates into just looking at whether $p$-divides the corner hook length or not.
Best Answer
Here is a solution using Pieri's rule:
The representation $\wedge^s V$ has as basis vectors: $$ \{e_{i_1}\wedge \dotsb \wedge e_{i_s}\mid 1\leq i_1<\dotsb <i_s\leq n\}. $$
If we restrict this representation to $S_{n-1}$, then the representation on the subspace spanned by $\{e_{i_1}\wedge \dotsb \wedge e_{i_s}\mid 1\leq i_1<\dotsb <i_s\leq n-1\}$ is just the representation of $S_{n-1}$ on $\wedge^s V_{n-1}$, where $V_{n-1}$ is the subspcace of $V$ consisting of vectors with the last coordinate equal to $0$.
On the other hand, the representation of $S_{n-1}$ on the subspace spanned by $\{e_{i_1}\wedge \dotsb \wedge e_{i_{s-1}}\wedge e_n\mid 1\leq i_1<\dotsb <i_{s-1}\leq n-1\}$ is isomorphic to the representation $\wedge^{s-1}V_{n-1}$ of $S_n$.
Therefore, by induction hypothesis, the restriction of $\wedge^s V$ to $S_{n-1}$ is the sum of the representation corresponding to $(n-s, 1^{s-1})$ and the representation corresponding to $(n-s-1, 1^s)$. It follows from Pieri's rule that $\wedge^s V$ is the representation corresponding to $(n-s, 1^s)$.