Find all solutions $(x_1, x_2, . . . , x_n)$ of the equation
\[1 +\frac{1}{x_1} + \frac{x_1+1}{x{}_1x{}_2}+\frac{(x_1+1)(x_2+1)}{x{}_1{}_2x{}_3} +\cdots + \frac{(x_1+1)(x_2+1) \cdots (x_{n-1}+1)}{x{}_1x{}_2\cdots x_n} =0\]
To solve the given equation, we start by analyzing the given expression:

\[1 +\frac{1}{x_1} + \frac{x_1+1}{x_1 x_2}+\frac{(x_1+1)(x_2+1)}{x_1 x_2 x_3} +\cdots + \frac{(x_1+1)(x_2+1) \cdots (x_{n-1}+1)}{x_1 x_2 \cdots x_n} = 0\]

1. **Rewrite the expression in a product form:**
   Notice that each term in the sum can be rewritten in a product form. Let's consider the general term:
   \[
   \frac{(x_1+1)(x_2+1) \cdots (x_{k-1}+1)}{x_1 x_2 \cdots x_k}
   \]
   This can be rewritten as:
   \[
   \prod_{i=1}^{k-1} (x_i + 1) \cdot \frac{1}{\prod_{i=1}^{k} x_i}
   \]

2. **Factor the entire expression:**
   We can factor the entire sum as follows:
   \[
   1 + \frac{1}{x_1} + \frac{x_1+1}{x_1 x_2} + \frac{(x_1+1)(x_2+1)}{x_1 x_2 x_3} + \cdots + \frac{(x_1+1)(x_2+1) \cdots (x_{n-1}+1)}{x_1 x_2 \cdots x_n}
   \]
   This can be written as:
   \[
   \prod_{i=1}^{n} \left(1 + \frac{1}{x_i}\right) - 1 = 0
   \]

3. **Simplify the equation:**
   Simplifying the above equation, we get:
   \[
   \prod_{i=1}^{n} \left(1 + \frac{1}{x_i}\right) = 1
   \]

4. **Analyze the product:**
   For the product of terms to equal 1, at least one of the terms must be zero (since the product of non-zero terms cannot be zero). Therefore, we need:
   \[
   1 + \frac{1}{x_j} = 0 \quad \text{for some } j
   \]
   Solving for \(x_j\), we get:
   \[
   \frac{1}{x_j} = -1 \implies x_j = -1
   \]

5. **Conclusion:**
   Therefore, the set of solutions \(E\) is:
   \[
   E = \{(x_1, x_2, \ldots, x_n) \mid x_i \neq 0 \text{ for all } i \text{ and at least one } x_j = -1\}
   \]

The final answer is \( \boxed{ E = \{(x_1, x_2, \ldots, x_n) \mid x_i \neq 0 \text{ for all } i \text{ and at least one } x_j = -1\} } \)