simplifying expressions - How to simplify exponents of the form $expleft(ifrac{2pi}{N}(N+1)right)$ inside a function

I'm really annoyed with Mathematica and I need your help. I defined a discrete Fourier-Transformation(I use II instead of N, because Mathematica wont let me-.-):

Subscript[ϕ, i_] = Sum[Exp[I (2 Pi)/II i k] f[k], {k, II}];

And I want to check that

Subscript[ϕ, 1] == Subscript[ϕ, II + 1]

This gives me $$\sum _k^{\text{II}} \left(f(k) e^{\frac{2 i \pi k}{\text{II}}}\right)=\sum _k^{\text{II}} \left(f(k) e^{\frac{2 i \pi (\text{II}+1) k}{\text{II}}}\right)$$

And by simply splitting the exponents on the right side of the equations this can be seen to be true.($\exp(i2\pi k)=1\forall k\in\mathbb N$ ) I really would have thought that Mathematica would be able to simplify that on its own, considering that $k$ is the summation index and is therefore an integer, but it does not. So I told Mathematica:

Simplify[Subscript[ϕ, 1] == Subscript[ϕ, II + 1],     Assumptions -> k ∈ Integers]

Didn't help, so I tried:

Simplify[Subscript[ϕ, 1] == Subscript[ϕ, II + 1] /. 
        Exp[a_] :> Exp[Expand[a]], Assumptions -> k ∈ Integers]

Still does not work! Please help me. Thank you

Answer

Apparently, simplifying a Sum does not result in simplifying each term in the Sum. Try

Subscript[ϕ, 1] == Subscript[ϕ, II + 1] /. 
    Exp[a_] :> Simplify[Exp[a], Assumptions -> k ∈ Integers]
(* True *)

Undoubtedly, there are more elegant approaches.