# Making Mathematica return the full simplified expression by default

I have an expression defined as

Ν[Δ_, l_] = FunctionExpand[((2^l (Δ + l - 1) Gamma[Δ + l - 1]^2 Gamma[Δ])/
(Gamma[Δ - 1] Gamma[(Δ + l)/2]^4 Gamma[1/8 - ((Δ - l)/2)]^2 *
Gamma[1/8 - ((2 - Δ - l)/2)]^2)),
Assumptions -> {Δ ∈ Integers, l ∈ Integers, Δ >= 0, l >= 0}];


The problem is that it diverges naively when I try to compute Ν[1,0] with the divergence coming from the fact that Mathematica evaluates Ν[k,0] to be

Ν[k, 0]
(* (2^(-4 + 2 k) (-1 + k)^2 Gamma[-(1/2) + k/2]^2)/(π Gamma[
1/8 - k/2]^2 Gamma[-(7/8) + k/2]^2 Gamma[k/2]^2) *)


Which means that the numerator becomes a product 0 ComplexInfinity. But when I use FullSimplify on the above, I get

Ν[k, 0] // FullSimplify
(* (4^(-1 + k) Gamma[(1 + k)/2]^2)/(π Gamma[
1/8 - k/2]^2 Gamma[-(7/8) + k/2]^2 Gamma[k/2]^2) *)


Is there a way to make Mathematica do this simplification by default? I am using this function in a sum, so I have to sum over $$l = 0$$ terms too, which is causing annoying errors.

f[kin_, Lin_] := Module[{num, den, L, k, t},
num = Limit[(2^(-4 + 3*L + 2*k)*(-1 + k)*(-1 + L + k)*
Gamma[-1/2 + L/2 + k/2]^2), {L -> Lin, k -> kin}];
den = Limit[(Pi*Gamma[1/8 + L/2 - k/2]^2*Gamma[-7/8 + L/2 + k/2]^2*
Gamma[L/2 + k/2]^2), {L -> Lin, k -> kin}];
t = num/den;
FunctionExpand[t,
Assumptions -> {k ∈ Integers, L ∈ Integers,
k >= 0, L >= 0}
]
]


And now

f[1, 0]
(* 0 *)


And

 f[k, 0] // FullSimplify