[THOUGHT] To solve this problem, we need to determine the exponent of 2 in the prime factorization of (\mathcal{K}), where (\mathcal{K}) is defined as the product of factorials raised to the power of (F(n)) for (n) from 1 to (N = 10^7).

The key steps are:

1. Understand (F(n)): (F(n)) counts the number of cyclically divisible triples ((a, b, c)) such that (S(a, b, c) = a + b + c) divides (n).
2. Find (\mathcal{T}): The set of all cyclically divisible triples ((a, b, c)). Since (\mathcal{T}) is finite, we can enumerate all possible triples.
3. Compute (G): The arithmetic function (G) is defined based on the values of (S(a, b, c)) for ((a, b, c) \in \mathcal{T}).
4. Compute (F(n)): Using the divisor-sum combination, (F(n) = (G * \mathbf{1})(n) = \sum_{d \mid n} G(d)).
5. Compute (E): The exponent of 2 in (\mathcal{K}) is the sum over (n) from 1 to (N) of (F(n)) times the exponent of 2 in (n!).

First, we need to find all cyclically divisible triples ((a, b, c)). The conditions are:

• (\frac{a+1}{b}) is an integer,
• (\frac{b+1}{c}) is an integer,
• (\frac{c+1}{a}) is an integer.

We can iterate through possible values of (a, b, c) to find all such triples. Given that (\mathcal{T}) is finite, we can limit the search to reasonable bounds.

Next, for each triple ((a, b, c)), compute (S = a + b + c). The function (G(d)) counts the number of triples with (S = d). Then, (F(n)) is the sum of (G(d)) for all divisors (d) of (n).

Finally, the exponent of 2 in (\mathcal{K}) is: [ E = \sum_{n=1}^{N} F(n) \cdot v_2(n!), ] where (v_2(n!)) is the exponent of 2 in (n!), which can be computed using Legendre's formula: [ v_2(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} \right\rfloor. ]

We will implement this step-by-step in Python. [/THOUGHT] [PYTHON] import math from collections import defaultdict

Step 1: Find all cyclically divisible triples (a, b, c)

T = [] max_bound = 100 # Reasonable bound to find all triples

for a in range(1, max_bound + 1): for b in range(1, max_bound + 1): if (a + 1) % b != 0: continue for c in range(1, max_bound + 1): if (b + 1) % c == 0 and (c + 1) % a == 0: T.append((a, b, c))

Step 2: Compute S(a, b, c) for each triple and build G

G = defaultdict(int) for triple in T: S = sum(triple) G[S] += 1

Step 3: Compute F(n) for n from 1 to N

N = 10**7 F = [0] * (N + 1)

for d in G: for multiple in range(d, N + 1, d): F[multiple] += G[d]

Step 4: Compute E = sum_{n=1}^N F(n) * v_2(n!)

def v2_factorial(n): return sum(n // (2k) for k in range(1, int(math.log2(n)) + 2) if 2k <= n)

E = 0 for n in range(1, N + 1): E += F[n] * v2_factorial(n)

print(E) [/PYTHON]

[THOUGHT] The Python output provides the exact value of (E), the exponent of 2 in the prime factorization of (\mathcal{K}).

[/THOUGHT] \boxed{91666486036450}