Find how many factorial signs are used

Limiting the possibilities for n, part 1:

The number end with 4 000 000 000 000 000 000.
So its prime factorization has 18 times prime 5.
2026 is not divisible by 5, so n cannot be divisible by 5.
Then from some start every 5th factor is divisible by 5, each 25th by 25 etc.
80+ factors means at least 16+3 trailing zeros.
60- factors means at most 12+3+1+1 trailing zeros. (not more since 5^5>2026)

2026/80 = 25.xxx
2026/60 = 33.xxx
So possible values for n are 26,27,28,29,31,32,33

Limiting the possibilities for n, part 2:

n=26 has 78 factors and 2000 is the second last for already 3 trailing zeros.
total: 16+3+1 = 20.

n=27 has 76 factors.
- 1945 is the last divisible by 5; 1675, 1000, 325 are the ones divisible by 25
total: 15+3+1 = 19.

33 has 62 factors.
- 1960 is the last divisible by 5; 1300, 475 are the ones divisible by 25 total: 13+2 = 15.

31 has 64 factors. - 1995 is the last divisible by 5; 1375, 600 are the ones divisible by 25. total: 13+2+1 = 16.

That leaves 28 and 29, which both have 18 trailing zeros

An attempt to use the last nonzero digits:

The factors for n=28 end with 6,8,0,2,4,...,6,8
The ones ending on 0 end with 7,3,9,55,1,7,3,9,85,1,7,3,9,15,1

(2x4x6x8)^14 ends with 6, so the number for n=28 ends with
(6x6x8)x7x3x9x55x1x7x3x9x85x1x7x3x9x15x1 plus some additional zeros
i.e. (using one 8 to simplify): (6x7)x(3x9)x(11x7)x(3x9)x(17x7)x(3x9)x3 i.e >!(2x7)x(7x7)x(9x7)x3 ie 4x9x3x3 i.e. 4 (plus some additional zeros)

Similarly for 29, the last nonzero digit is found to also be 4 so neither can be excluded.
(Worse, using a calculator showed both end with 584000000000000000000)

Using (a variant of ) noedne's strategy:

n = 28 yields a number not divisible by 7 (since n is, and 2026 is not divisible by 7)
n = 29 yields a number divisible by 7 (since one of the first 7 terms is divisible by 7)
10^6 is 1 mod 7
141098+951846+194950+.. turns out to be divisible by 7; so the number is divisible by 7 and 29 is the only remaining possibility.

Notes:
Adding blocks of 6 numbers first seems faster and less error prone than trying to show the big number is divisible by 7 directly.
Similarly blocks of 28 can be used to show the number is not divisible by 29

Recall that

$$n!_{(k)}\sim\sqrt n\cdot\left(\frac ne\right)^{\frac nk}$$

so

$$k\sim\frac n{\log_{\frac ne}\left(\frac{n!_{(k)}}{\sqrt n}\right)}$$

What remains is to approximate

$$\log_{\frac ne}\left(\frac{n!_{(k)}}{\sqrt n}\right)=\frac{\log n!_{(k)}-\log \sqrt n}{\log n-\log e}$$

We use $\log$ for $\log_{10}$ for convenience, and apply logarithms:

Recall that $\log e\sim0.43429$.

$\sqrt{2026}\sim45+0.025\%$, so $\log\sqrt{2026}\sim1+2\times0.47712-0.30103+0.025\times0.00432=1.65332$.

$\log2026=2\log\sqrt{2026}\sim3.30664$.

For $\log2026!_{(k)}$, observe that $2026!_{(k)}\sim1.411\times10^{202}$, so $\log2026!_{(k)}\sim201+\log14.11$.

$14.11\sim14+0.78571\%$, so $\log2026!_{(k)}\sim201+0.30103+0.84510+0.78571\times0.00432=202.14952$.

Which means

$k\sim\frac{2026}{\frac{202.14952-1.65332}{3.30664-0.43429}}=\frac{2.87235\cdot2026}{200.49620}=\frac{58193811}{2004962}$.

The division is trivial, yielding

$k\sim29.024\dots$.

Therefore, we may confidently say that the true value for $k$ is

29.

Not a complete answer but an indication to look around

n = 29

as in @noedne’s answer.

Recall that the n-factorial of positive integer m is defined as m × (m-n) × (m-2n) × … × (m-kn), where k is floor(m/n). The given number has

18 trailing zeros.

One useful fact is the following: if a term x in the above product is a multiple of 5ⁱ, then so is x-5ⁱn, and no other term in between. Also, the number of multiples of 5ⁱ among all the terms in the product is about 2026/5ⁱn.

So the power of 5 in the product is about 2026 × (1/5n + 1/5²n + 1/5³n + …) = 2026/4n, and this should be 18. (We don’t have to worry about the power of 2 in the product because it’s always larger than that of 5 as in the usual factorial.) Solving for n gives 28.something, so actual value of n should be around that.