Proof of the inclusion-exclusion principle The inclusion-exclusion principle for $n$ sets is proved by Kenneth Rosen in his textbook on discrete mathematics as follows:


  THEOREM 1 &mdash; THE PRINCIPLE OF INCLUSION-EXCLUSION &nbsp; Let $A_1,A_2,\ldots,A_n$ be finite sets. Then
  $$
\begin{multline*}
|A_1 \cup A_2 \cup \cdots \cup A_n| = \sum_{1 \leq i \leq n} |A_i| - \sum_{1 \leq i &lt; j \leq n} |A_i \cap A_j| \\ + \sum_{1 \leq i &lt; j &lt; k \leq n} |A_i \cap A_j \cap A_k| - \cdots + (-1)^{n+1} |A_1 \cap A_2 \cap \cdots \cap A_n|.
\end{multline*}
$$
  Proof: We will prove the formula by showing that an element in the union is counted exactly once by the right-hand side of the equation. Suppose that $a$ is a member of exactly $r$ of the sets $A_1,A_2,\ldots,A_n$, where $1 \leq r \leq n$. This element is counted $\binom{r}{1}$ times by $\sum |A_i|$. It is counted $\binom{r}{2}$ times by $\sum |A_i \cap A_j|$. In general, it is counted $\binom{r}{m}$ times by the summation involving $m$ of the sets $A_i$. Thus, this element is counted exactly
  $$
\binom{r}{1} - \binom{r}{2} + \binom{r}{3} - \cdots + (-1)^{r+1} \binom{r}{r}
$$
  times by the expression on the right-hand side of this equation. Our goal is to evaluate this quantity. The binomial formula shows that
  $$
0 = (1-1)^r = \binom{r}{0} - \binom{r}{1} + \binom{r}{2} - \cdots + (-1)^r \binom{r}{r}.
$$
  Hence
  $$
1 = \binom{r}{0} = \binom{r}{1} - \binom{r}{2} + \binom{r}{3} - \cdots + (-1)^{r+1} \binom{r}{r}.
$$
  Therefore, each element in the union is counted exactly once by the expression on the right-hand side of the equation. This proves the principle of inclusion-exclusion.


Although the proof seems very exciting, I am confused because what the author has proved is $1=1$ from the LHS and RHS.

Thus, is this still a valid proof? We need to prove that the total cardinality of LHS is the RHS. The RHS produces a $1$ for each member of the union of the sets. 

I think in order to produce the cardinality of the union, an extra
summation sign should be appended before the expression in RHS. Could
someone please clarify?
Let me slightly rephrase the argument. Let $N_r$ be the number of elements contained in exactly $r$ of the sets $A_1,\ldots,A_n$. Then the left-hand side is
$$
|A_1 \cup \cdots \cup A_n| = \sum_{r=1}^n N_r.
$$
The first sum on the right-hand side is
$$
\sum_{1 \leq i \leq n} |A_i| = \sum_{r=1}^n \binom{r}{1} N_r.
$$
The second sum on the right-hand side is
$$
\sum_{1 \leq i &lt; j \leq n} |A_i \cap A_j| = \sum_{r=2}^n \binom{r}{2} N_r.
$$
More generally,  the $m$th sum on the right-hand side is
$$
\sum_{1 \leq i_1 &lt; i_2 &lt; \cdots &lt; i_m \leq n} |A_{i_1} \cap A_{i_2} \cap \cdots \cap A_{i_m}| = \sum_{r=m}^n \binom{r}{m} N_r.
$$
Therefore the right-hand side is equal to
$$
\sum_{r=1}^n \binom{r}{1} N_r - \sum_{r=2}^n \binom{r}{2} N_r + \sum_{r=3}^n \binom{r}{3} N_r - \cdots + (-1)^{n+1} \sum_{r=n}^n \binom{r}{n} N_r.
$$
Rearranging, the right-hand side is equal to
$$
\left[ \binom{1}{1} \right] N_1 + \left[ \binom{2}{1} - \binom{2}{2} \right] N_2 + \left[ \binom{3}{1} - \binom{3}{2} + \binom{3}{3} \right] N_3 + \cdots + \\
\left[ \binom{n}{1} - \binom{n}{2} + \binom{n}{3} - \cdots + (-1)^{n+1} \binom{n}{n} \right] N_n.
$$
The coefficient of the general term $N_m$ is
$$
\binom{m}{1} - \binom{m}{2} + \binom{m}{3} - \cdots + (-1)^{m+1} \binom{m}{m}.
$$
By the binomial theorem, this equals $1$, and so the right-hand side equals
$$
1 \cdot N_1 + 1 \cdot N_2 + 1 \cdot N_3 + \cdots + 1 \cdot N_n = \sum_{r=1}^n N_r,
$$
which is exactly the same as the left-hand side.