I want to complete the following proof:
So continuing where the author left off, I did the following:
$\begin{array} {cc} & \sum\limits_{i=1}^{n-1} P(A_i) - \sum\limits_{1\le i < j \le n-1}P(A_i \cap A_j) + \dots + (-1)^n P(A_1 \cap A_2 \cap \dots \cap A_{n-1}) + P(A_n) - P(\bigcup\limits_{i=1}^{n-1} (A_i \cap A_n))\\ &= \sum\limits_{i=1}^{n-1} P(A_i) - \sum\limits_{1\le i < j \le n-1}P(A_i \cap A_j) + \dots + (-1)^n P(A_1 \cap A_2 \cap \dots \cap A_{n-1}) + P(A_n) - \left[ \sum\limits_{i=1}^{n-1}P(A_i \cap A_n) - \sum\limits_{1\leq i < j \leq n-1}^{n-1}P(A_i \cap A_j \cap A_n) + \dots + (-1)^nP(A_i \cap A_2 \cap \dots \cap A_{n-1}\cap A_n)\right]\\ \end{array}$
I got here by applying the inclusion exclusion formula to $P(\bigcup\limits_{i=1}^{n-1} (A_i \cap A_n)$ as the author instructed and then simplyfying based on the following:
$(A_i \cap A_n) \cap (A_j \cap A_n) = A_i \cap A_j \cap A_n$
I tried this new formula with $n=3$ and I did get the correct answer, so I think that the manipulations I did is correct.
Also, I know I can get the $\sum\limits_{i=1}^{n}P(A_i)$ in the theorem because:
$\sum\limits_{i=1}^{n}P(A_i) = \sum\limits_{i=1}^{n-1}P(A_i) + P(A_n)$
So, this formula can be simplified to:
$\sum\limits_{i=1}^{n} P(A_i) - \sum\limits_{1\le i < j \le n-1}P(A_i \cap A_j) + \dots + (-1)^n P(A_1 \cap A_2 \cap \dots \cap A_{n-1}) - \left[ \sum\limits_{i=1}^{n-1}P(A_i \cap A_n) - \sum\limits_{1\leq i < j \leq n-1}^{n-1}P(A_i \cap A_j \cap A_n) + \dots + (-1)^nP(A_i \cap A_2 \cap \dots \cap A_{n-1}\cap A_n)\right]$
Now I don't know how to proceed.
If the answer involves manipulating the summations then the more detailed the answer, the better as I am not too familiar with manipulating sums.