Proving the addition theorem for probabilities using induction Given $A_1,...,A_n$ in a discrete probability space, I have to prove:$$P\left(\bigcup\limits_{k=1}^{n}A_k\right)=\sum \limits_{\varnothing \neq J\subseteq \{1,2,...,n\}}(-1)^{|J|+1}P\left(\bigcap\limits_{j\in J}A_j\right)$$ using induction.
My approach
base case
$$P\left(\bigcup\limits_{k=1}^{1}A_k\right)=P(A_1)=P(A_1)=\sum \limits_{\varnothing \neq J\subseteq \{1\}}(-1)^{|J|+1}P\left(\bigcap\limits_{j\in J}A_j\right) \quad \checkmark$$
inductive step
$$P\left(\bigcup\limits_{k=1}^{n+1}A_k\right)= P\left(\left(\bigcup\limits_{k=1}^{n}A_k\right)\cup A_{n+1}\right)= P\left(\bigcup\limits_{k=1}^{n}A_k\right)+P(A_{n+1})-P\left(\left(\bigcup\limits_{k=1}^{n}A_k\right)\cap A_{n+1}\right) \\
=\left(\sum \limits_{\varnothing \neq J\subseteq \{1,2,...,n\}}(-1)^{|J|+1}P\left(\bigcap\limits_{j\in J}A_j\right)\right)+\underbrace{P(A_{n+1})-P\left(\left(\bigcup\limits_{k=1}^{n}A_k\right)\cap A_{n+1}\right)}_{\text{This should equal the } (n+1) \text{ summand}}$$
The $n+1$ summand is: $(-1)^nP(A_1\cap A_2\cap \cdots \cap A_{n+1})$.
Hence the proof breaks down to showing $$P(A_{n+1})-P\left(\left(\bigcup\limits_{k=1}^{n}A_k\right)\cap A_{n+1}\right)=(-1)^nP(A_1\cap A_2\cap \cdots \cap A_{n+1})$$
That's where I'm stuck. Can anyone help?
 A: The reason you're stuck is that you've made a mistake leading to the next result that you're trying to show.
Starting from this line:

$$= \left(\sum \limits_{\varnothing \neq J\subseteq \{1,2,...,n\}}(-1)^{|J|+1}P\left(\bigcap\limits_{j\in J}A_j\right)\right)+\underbrace{P(A_{n+1})-P\left(\left(\bigcup\limits_{k=1}^{n}A_k\right)\cap A_{n+1}\right)}_{\text{This should equal the } (n+1) \text{ summand}}$$

I think there's more missing here than you might be seeing -- notice that you hope to show that this entire expression is equal to
$$\left(\sum \limits_{\varnothing \neq J\subseteq \{1,2,...,\color{blue}{\fbox{n+1}}\}}(-1)^{|J|+1}P\left(\bigcap\limits_{j\in J}A_j\right)\right)$$
which means that you're not just missing "the $(n+1)$ summand" -- you're missing all intersections that include $A_{n+1}$. That is, there are terms corresponding to $J = \{1, n+1\}$, and $\{2, n+1\}$, etc. -- and $J = \{1, 2, n+1\}$, and so forth. For this reason, you can't show the result you're trying to show; those expressions aren't equal.
As for how to proceed; consider taking your underbraced term and rewriting
$$\left( \left( \bigcup_{k=1}^n A_k \right) \cap A_{n+1} \right) =  \bigcup_{k=1}^n \left( A_k \cap A_{n+1} \right) $$
to proceed. (At least, I think that should work out.)
