Another way to get at this:

  φ^1 = φ
  φ^2 = φ + 1 [this is the definition of φ]
  φ^3 = φ(φ + 1) = 2φ + 1
  φ^4 = φ(2φ + 1) = 3φ + 2
  φ^5 = φ(3φ + 2) = 5φ + 3
  φ^6 = φ(5φ + 3) = 8φ + 5
  ...
  φ^n = F(n)φ + F(n-1)

Note that F(n+1)/F(n) → φ as n → ∞. Or flipped around, F(n)φ → F(n+1).

So φ^n = F(n)φ + F(n-1) → F(n+1) + F(n-1) = L(n) as n → ∞. This gets close to being an integer because F(n+1) and F(n-1) are both integers.