Using Inclusion-Exclusion, probability of occurrence of exactly $m$ events out of $\{A_1,\ldots,A_n\}$ Using Inclusion-Exclusion, prove that the probability of occurrence of exactly $m$ events out of $\{A_1,\ldots,A_n\} = \sum_{i=0}^{n-m} (-1)^i   \binom{m+i}{m} P_{m+i} $ where $P_k = \sum_{1 \leq i_1<\cdots \le i_k \leq n} P(A_{i_1} \cap\cdots \cap A_{i_k})$
Attempt:  I was trying to understand this term on the RHS : $\binom{m+i}{m} P_{m+i}$. And then the inclusion-exclusion principle shall follow.
This is what I thought to be able to use inclusion-exclusion but I am not very sure if this is the right approach.
$m$ events can occur either from the first $m$ events, or from the first $m+1$ events, or from the first $m+2$ events, $\cdots$, or from the n events.
$$E_k = \text{the $m$ events occur from the first $m+k$ events}$$
Then $P(E_k) = \binom{m+k}{m} \sum_{1\leq i_1<\cdots i_m\leq m+k} P(A_{i_1} \cap \cdots  \cap A_{i_m})$
But, then I could not move forward from here. I tried understanding the solution given here, but I would really appreciate something more intuitive.
Could someone please explain how do I move forward from here?

I discovered a marvelous proof. However, since it's long, I haven't been able to type it. Posting clear images for the sake of anyone who might be searching for the same answer.
