Consider the following functions: f₁(n) = 2n f₂(n) = nn f₃(n) = n! f₄(n) =…

Key idea: Compare the functions pairwise to determine which grows faster as n → ∞.

• Compare n^n and n!: Consider the ratio n^n / n! = ∏_{k=1}^n (n/k). Many factors in this product are ≥2 (for k ≤ n/2), so the product grows without bound. Therefore n^n grows faster than n!.

• Compare n! and 2^n: By Stirling's approximation, n! ≈ (n/e)^n √(2πn). Hence n!/2^n ≈ (n/(2e))^n √(2πn) → ∞, so n! grows faster than 2^n.

• Compare 2^n and n log log n: An exponential function like 2^n eventually dominates any function that is polynomial or near-linear in n (including n·log log n). For example, taking logarithms gives n·log 2 versus log n + log(log log n), and n·log 2 grows without bound compared to log n, so 2^n≫n·log log n.

Final decreasing order: n^n > n! > 2^n > n·log log n

In terms of the given functions: f2, f3, f1, f4.