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)

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.

To show what we want, we must prove the stronger condition that F(n)φ - F(n+1) → 0. Merely knowing that F(n+1)/F(n) → φ only shows that the difference grows more slowly than F(n).

Noting that {F(n+1)/F(n)} is the sequence of convergents of the continued fraction expansion of φ, your stronger condition follows from Theorem 5 at https://en.wikipedia.org/wiki/Continued_fraction .

Given a polynomial P(x), any expression that can be written as a symmetric polynomial in the roots of P, can be written as a polynomial in the coefficients of P. See https://en.wikipedia.org/wiki/Elementary_symmetric_polynomia... for more.

This applies to arbitrary rings. So if P(x) and Q(x) are both polynomials, then any polynomial which is symmetric in the roots of P(x) and also symmetric in the roots of Q(x) can be written as polynomials in the coefficients of Q(x) and P(x). Therefore, for example, (x-sqrt(2)-sqrt(3))(x-sqrt(2)+sqrt(3))(x+sqrt(2)-sqrt(3))(x+sqrt(2)+sqrt(3)) will work out to be a polynomial in the coefficients of x^2-2 and x^2-3. It will therefore be an integer polynomial with those 4 roots.

This is actually the original way in which people proved that the algebraic integers formed a ring.

This is the same fact twice. (Well, the second part is just one of the two roots of the polynomial equation x^2 = x + 1.)

L_{n-1} = 0 ;-)

You probably meant L_n = L_{n-2} + L_n{n-1}