Find the number of integer solutions of the equation
$x^{2016} + (2016! + 1!) x^{2015} + (2015! + 2!) x^{2014} + ... + (1! + 2016!) = 0$
1. **Consider the given polynomial equation:**
   \[
   P(x) = x^{2016} + (2016! + 1!) x^{2015} + (2015! + 2!) x^{2014} + \cdots + (1! + 2016!) = 0
   \]

2. **Analyze the polynomial modulo 2:**
   - If \( x \) is odd, then \( x \equiv 1 \pmod{2} \).
     \[
     x^{2016} \equiv 1^{2016} \equiv 1 \pmod{2}
     \]
     Each term \( (k! + (2017-k)!) x^{2016-k} \) for \( k = 1, 2, \ldots, 2016 \) can be analyzed modulo 2:
     \[
     k! \text{ and } (2017-k)! \text{ are both even for } k \geq 2 \implies k! + (2017-k)! \equiv 0 + 0 \equiv 0 \pmod{2}
     \]
     For \( k = 1 \):
     \[
     1! + 2016! \equiv 1 + 0 \equiv 1 \pmod{2}
     \]
     Therefore, the polynomial modulo 2 when \( x \) is odd:
     \[
     P(x) \equiv 1 + 0 + 0 + \cdots + 1 \equiv 1 \pmod{2}
     \]
     This implies \( P(x) \) is odd when \( x \) is odd.

3. **Analyze the polynomial for even \( x \):**
   - If \( x \) is even, then \( x \equiv 0 \pmod{2} \).
     \[
     x^{2016} \equiv 0^{2016} \equiv 0 \pmod{2}
     \]
     Each term \( (k! + (2017-k)!) x^{2016-k} \) for \( k = 1, 2, \ldots, 2016 \) can be analyzed modulo 2:
     \[
     k! + (2017-k)! \equiv 0 \pmod{2} \text{ for } k \geq 2
     \]
     For \( k = 1 \):
     \[
     1! + 2016! \equiv 1 + 0 \equiv 1 \pmod{2}
     \]
     Therefore, the polynomial modulo 2 when \( x \) is even:
     \[
     P(x) \equiv 0 + 0 + 0 + \cdots + 1 \equiv 1 \pmod{2}
     \]
     This implies \( P(x) \) is odd when \( x \) is even.

4. **Conclusion:**
   Since \( P(x) \) is always odd regardless of whether \( x \) is odd or even, it can never be zero. Therefore, there are no integer solutions to the equation.

\(\blacksquare\)

The final answer is \( \boxed{ 0 } \) integer solutions.