We denote 100! as the integer $1 \times 2 \times 3 \times \cdots \times 100$. Determine the smallest integer strictly greater than 100 that is a divisor of $100!$.
Answer: 102. Since $102=2 \times 51$, it divides 100!. However, 101 is a prime number and does not divide any of the factors of $1 \times 2 \times \cdots \times 100$, so it does not divide 100!.