This is one possible solution path.
In several cells the number of digits can be determined immediately.
For example, 33ac is a six-digit number in three cells,
so it must be entered as *nn*|*nn*|*nn*
(using vertical bars to show cell divisions),
which means 24dn and 25dn are *n*|*n*|*nn*
and 17dn is *n*|*n*|*n*|*nn*,
forcing 25ac to be *n*|*n*|*nn*,
26dn to be *nn*|*n*|*n* and so on.
Because the cell contents must eventually be
replaced by letters representing 1 to 19,
single-digit cells can only contain 1 to 9 (not 0)
and two-digit cells must contain 10 to 19.

Rearranging 21ac,
it’s (R^{R} + R) + (TA + IE) − S and has four digits.
The maximum for (TA + IE) is (16×17 + 18×19) = 614.
If R is 4 then (4^{4} + 4) + 614 = 874 is too small;
if R is 6 then (6^{6} + 6) = 46662 is too big;
therefore **R = 5**.

For 14ac,
the factorial T! has no more than four digits,
so T ≤ 7 (7! = 5040, 8! = 40320).
For 30ac,
W^{(A − S)} − O + K has three digits,
so (A − S) ≥ 2, which means A ≥ 3.
From 22dn,
(R − A) > 0, so A ≤ 4.
Then for 29dn = T^{A} to have three digits the possibilities are
6^{3} = 216 and 7^{3} = 343
(not 4^{4} because T and A must be different,
and not 5^{3} or 5^{4} because 5 is already taken);
but 29dn is a palindrome,
so **T = 7** and **A = 3**
and we can enter **29dn = 343** with one digit per cell.
To make (A − S) ≥ 2,
we must have **S = 1**.

1ac is now (15M)^{C} + H,
with five digits.
If C is 4,
the minimum possible is (15×2)^{4} + 6 = 810006,
too big,
so C ≤ 3; but 1 and 3 are already taken,
so **C = 2**.
27dn is now 2^{E}(4 + M + I) − 1 and has four digits,
entered in three cells.
The minimum for 27dn is 10|1|1,
so the minimum for 2^{E}
is (1011 + 1)/(4 + 19 + 18) ≈ 24.7,
so E ≥ 6
(2^{4} = 16, 2^{5} = 32,
but 5 is already taken).
For 31ac,
the factorial E! must have no more than three digits,
so E ≤ 6,
which forces **E = 6**.

21ac is now 3150 + 6I,
in the range 3150 + 6×4 = 3174 to 3150 + 6×19 = 3264.
To fit into three cells, each containing 1 to 19,
it must be either 3|17|1 to 3|19|9, or 3|2|10 to 3|2|19,
which makes 22dn start with 17, 18, 19 or 2 in the first cell.
22dn is 2B^{3} − 7 and has four digits,
so B is between ((1711 + 7)/2)^{(1/3)} ≈ 9.5
and ((2919 + 7)/2)^{(1/3)} ≈ 11.4,
ie one of {10, 11}.
But 2×11^{3} − 7 = 2655
can’t be partitioned into three cells,
so **B = 10** and **22dn = 1993**,
entered as 19|9|3.
21ac is now in the range 3|19|1 to 3|19|9,
and 3150 + 6I = 6(525 + I) is a multiple of both 2 and 3,
which allows only {3192, 3198},
so I is one of {7, 8}; but 7 is taken,
so **I = 8** and **21ac = 3198**.

In 33ac,
the E^{(I − G)} term is now 6^{(8 − G)},
so to avoid a fractional value we need G < 8
and the only available value is **G = 4**.
The clue is then 10^{(O − 10)}L + 1314 + 7H + M,
with six digits in three cells,
so 10^{(O − 10)} is between
(101010 − 1314 − 7×18 − 17)/19 ≈ 5239.6 and
(191919 − 1314 − 7×11 − 12)/9 ≈ 21168.4;
the only power of 10 in that range is 10^{4} = 10000,
so **O = 14**.
Now 31ac = 392L − 1, to fit 43_ in the grid,
with two digits in the last cell;
392×9 − 1 = 3527 is too low and
392×12 − 1 = 4703 is too high,
so **L = 11** and **31ac = 4311**.
That also completes **26dn = 1856**,
which we can enter as 18|5|6.

34ac = 8N^{2} + 5 and has three digits,
so N < √((1000 − 5)/8) ≈ 11.2;
the only available value is **N = 9**,
giving **34ac = 653**,
**25ac = 7|3|18**,
**14ac = 5|15|1**,
**3dn = 7|11|1|18**,
**1dn = 8|8|14|6**,
and **7ac = 19|6|5**.
For 7dn we have 53M^{2} to fit 19_33 in the grid,
so M^{2} ends in 1 and M ends in 1 or 9,
for which the only available value is **M = 19**,
making **7dn = 19|1|3|3**,
**9dn = 5|13|1**,
**4ac = 6|9|16**,
**10ac = 823**,
**11dn = 2|5|19**,
**18ac = 18|8|9|4**,
**27dn = 19|8|3** and **35ac = 4389**.

In the grid we have 1|_|13 for 13ac,
so 8dn is 6__ with two digits in the last cell.
It’s a palindrome,
so the last cell must contain 16,
ie **8dn = 6|1|16**; its clue is 44(123 + U),
so **U = 16**,
which makes **13ac = 1|1|13**,
**5dn = 9569**,
**6dn = 16|14|2**,
**16ac = 6|19|3** and **24dn = 1|1|11**.

The grid now has **15ac = 2316** and its clue is 8V^{2} + 4,
so **V = 17**,
which gives **17dn = 3|3|3|17**,
**19dn = 8675** and **23dn = 8|1|11|9**.

1ac is 81225 + H,
to fit 8__7 in the grid,
so H ends in 2 and can only be **H = 12**,
making **1ac = 8|12|3|7** and **33ac = 11|14|17**.
The grid now has **2dn = 3|3|15**,
for which the clue is 255W,
so **W = 13** and
the only remaining variable is **D = 15**.
This completes the grid with **12ac = 11|2|5**,
**4dn = 6|2|10**,
**30ac = 173**,
**25dn = 7|7|14**,
**31ac = 789** and **20dn = 4694**.

Finally, changing each cell’s contents to the corresponding letter spells out the message I HATE NUMERICAL CROSSWORDS BECAUSE MAKING A MISTAKE MEANS STARTING ALL OVER AGAIN.