What is the hardest math problem ever solved?

This is a fantastic question because it exposes a deep truth about how exponents work. **The pattern method:** Look at powers of 2: 2^4 = 16, 2^3 = 8, 2^2 = 4, 2^1 = 2. Notice the pattern — each time you decrease the exponent by 1, you divide by 2. Following this pattern: 2^0 = 2 / 2 = 1. This works for any number: 3^0 = 1, 100^0 = 1, (anything)^0 = 1. **The algebraic reason:** By definition, x^a × x^b = x^(a+b). So x^n × x^0 = x^(n+0) = x^n. This means x^n × x^0 = x^n. Divide both sides by x^n (assuming x ≠ 0), and we get x^0 = 1. It's the only value that makes the exponent rules consistent. **Why not zero?** If x^0 = 0, then x^n × 0 = 0, but x^(n+0) = x^n ≠ 0. The rules would break. So x^0 must be 1 to keep the entire system of exponents logically consistent. **The exception:** 0^0 is a special case that's often defined as 1 in combinatorics and discrete math, but is left undefined in analysis. It depends on context.