General inclusion–exclusion

Fix any element $x$. If $x$ belongs to none of $A_1,\ldots,A_n$, it contributes $0$ to both sides of the formula, so assume $x$ belongs to exactly $m\ge 1$ of the sets.

For each $k$, the number of $k$-fold intersections $A_{i_1}\cap\cdots\cap A_{i_k}$ that contain $x$ equals the number of ways to choose $k$ of the $m$ sets containing $x$, namely $\binom{m}{k}$. So $x$'s total contribution to the right-hand side is $\sum_{k=1}^n(-1)^{k+1}\binom{m}{k}=\sum_{k=1}^m(-1)^{k+1}\binom{m}{k}$ (terms with $k>m$ vanish since $\binom{m}{k}=0$).

By the binomial theorem, $\sum_{k=0}^m(-1)^k\binom{m}{k}=(1-1)^m=0$, so $\sum_{k=1}^m(-1)^k\binom{m}{k}=-1$, and multiplying by $-1$ gives exactly $\sum_{k=1}^m(-1)^{k+1}\binom{m}{k}=1$.

So every element in at least one set contributes exactly $1$ to the right-hand side, matching its contribution of $1$ to $\left|\bigcup_{i=1}^n A_i\right|$; elements in no set contribute $0$ to both sides. Since every element contributes equally to both sides, the two sides are equal.