This is one possible solving path.
We’ll use subscripts X_{a} and X_{d}
to refer to the respective across and down values of X.
It will be useful to compile a list of the first factorial values,
{1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720,
7! = 5040, 8! = 40320, 9! = 362880, …}.

From 12ac, which has two digits, we have (using across values throughout)
O! = 12ac − A − T − M − C + I,
so O! ≤ 99 − 2 − 4 − 6 − 8 + 33 = 112.
Since O ≥ 2, O! must be one of {2, 6, 24}, with O in {2, 3, 4}.
For 11ac to be positive we need O > N,
so N = 2|3 (not yet knowing which is across or down) with
**O _{a} = 4** and

11ac is a factorial and must be 8! = 40320 (the only one with five digits),
so (P_{a} + H_{a})(4 − N_{a})/4^{Na} = 8,
which means P_{a} + H_{a} =
8×4^{Na}/(4 − N_{a}).
If N_{a} is 3 that gives 512,
but the maximum sum for any two distinct variables is 31 + 33 = 64;
so **N _{a} = 2**,

33ac is a three-digit factorial, one of {5! = 120, 6! = 720},
so P_{a} − I_{a} is one of {13, 14};
the two possibilities for P_{a} mean
I_{a} is one of {17, 18, 19, 20}
and I_{d} is in {16, 19, 18, 21} respectively.
8dn is a cube matching _2_, ie one of {5^{3} = 125, 9^{3} = 729},
so S_{d}!/P_{d} − I_{d} is in {5, 9}.
The lower limit for S_{d}! is 30(5 + 16) = 630
and the upper limit is 32(9 + 21) = 960;
the only factorial in that range is 720,
so **S _{d} = 6** and

Now for 6ac we have 7808 + R_{a} = __17,
which can only work if **R _{a} = 9**
and

28dn is now 150(41 − T_{d}), which clearly ends in 0.
That makes 90 = 41ac = B_{a} + 77,
so **B _{a} = 13** and

For 37ac = 10829 + M_{a} we have __8_7 in the grid,
so M_{a} ends in 8 and the only available value is **M _{a} = 28**,
with

In the grid we have 1_1_ for 18dn = 8(E_{d} + 117), which must be even.
Thus 31ac = E_{a} + 2 ≥ 21;
the only available values that make that work are {22, 23},
so 18dn ends in 2, E_{d} + 117 ends in 4 or 9,
and E_{d} ends in 7 or 2.
Thus **E _{d} = 22**,

Now we have 38 in the grid for 4ac = 53 − A_{a},
so **A _{a} = 15** and

The only remaining unassigned variable is C = 16|17.
We have 92 in the grid for 12ac = C_{a} + 75,
so **C _{a} = 17** and