Let $a \star b = ab + a + b$ for all integers $a$ and $b$.  Evaluate $1 \star ( 2 \star ( 3 \star (4 \star \ldots ( 99 \star 100 ) \ldots )))$.
1. Given the operation \( a \star b = ab + a + b \), we can rewrite it in a more convenient form. Notice that:
   \[
   a \star b = ab + a + b = (a+1)(b+1) - 1
   \]
   This can be verified by expanding the right-hand side:
   \[
   (a+1)(b+1) - 1 = ab + a + b + 1 - 1 = ab + a + b
   \]
   Thus, the given operation can be expressed as:
   \[
   a \star b = (a+1)(b+1) - 1
   \]

2. We need to evaluate \( 1 \star (2 \star (3 \star (4 \star \ldots (99 \star 100) \ldots ))) \). Using the rewritten form, we can simplify the nested operations step by step.

3. Start with the innermost operation:
   \[
   99 \star 100 = (99+1)(100+1) - 1 = 100 \cdot 101 - 1 = 10100 - 1 = 10099
   \]

4. Next, consider the next operation:
   \[
   98 \star 10099 = (98+1)(10099+1) - 1 = 99 \cdot 10100 - 1 = 999900 - 1 = 999899
   \]

5. Continue this process iteratively. Notice the pattern:
   \[
   a \star (b \star c) = a \star ((b+1)(c+1) - 1) = (a+1)((b+1)(c+1)) - 1
   \]
   This pattern generalizes to:
   \[
   1 \star (2 \star (3 \star (4 \star \ldots (99 \star 100) \ldots ))) = (1+1)(2+1)(3+1)\ldots(99+1)(100+1) - 1
   \]

6. Therefore, the expression simplifies to:
   \[
   (1+1)(2+1)(3+1)\ldots(99+1)(100+1) - 1 = 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 100 \cdot 101 - 1
   \]

7. Recognize that the product \( 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 100 \cdot 101 \) is the factorial of 101:
   \[
   101! - 1
   \]

Conclusion:
\[
\boxed{101! - 1}
\]