Vtyjkyuk

Probably not the intended answer, but, I propose:

$108+sqrtsqrtldots sqrt2$, with infinitely many square roots.

Explanation:

Formally:
$$sqrtsqrtldots sqrt2=lim_limitsnto infty s_n=1$$

where $s_0=2$ and $s_n+1=sqrts_n$.

This is technically a solution with n-druple factorials. It is not a good solution.

Solution:

$$frac(10!!!)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!(2+8)!!!.$$

Explanation:

First, we note that $2+8=10$. In particular, we can construct the number $10cdot 7cdot 4cdot 1=10!!!=280$ in two different ways.
Once we have two copies of $280$, we can construct the number $$280!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!$$ (that's $171$ factorials, so it equals $280cdot 109$), and then simply divide by $280$ again. This technique can be used to boringly nuke every problem of this form that allows for n-druple factorials: if you want to get some number $K$, you can use the property that $2+8=10$, so you can get two copies of $N$ for some large $N$ (by repeatedly taking factorials from $10$, for example; in terms of big $N$, any $Ngeq 2K$ should do it), and then you can take $N!_N-K$ ($N-K$ factorials) to get the number $Ncdot K$ - then you just divide by $N$ and now you have $K$.

If you allow

decimals

then you can do

$$109=(.1)^-2 + 8 + 0!$$

The trick is to:

... count in hexadecimal: 21 × 8 + 0! = 109


(in decimal: 21h = 33 and 109h = 265)

Hmmm, possibly:

Place a vertical mirror by the $2$ to get a $5$. Concatenate the $5$ with the $0$ to get $50$ and take the $38$-factorial (using $38$ exclamations): $50!^38=50cdot12=600$, and then add $1^8=1$: $50!^38+1^8=601$. Turn the $601$ upside-down to get $109$.

and for posterity:

$20!^15+1+8$

This probably won't count. It uses $!$, but not in $x!$.

$$108 + !2$$

Explanation

$!n$ is the number of derangements of n objects. In particular $!2 = 1$.

I have a few silly answers :) Though, frustratingly, I haven't been able to solve it yet

$108$++ $= 109$ which you might also write as $108$+^$2$

or if we are allowing transformation of numbers as I've seen above

$(2+8)$ concatenated with $0$, with a vertical line (using the $1$) on the RHS to turn it into a $9$

or

$8-2=6$, which rotated gives $9$, then concatenate $10$ to give $109$

I was thinking about trying to

Change the base of the numbers

but didn't get anywhere with that idea...

Looking forward to seeing the solution!

$$8+2^0=9=09implies 18+2^0=109$$

I break the rules, BUT!

work with 1 and 20
1 - 20 = -19

109 = arccos(sin(-19))

I have another trigonometric answer.

$$8^2 - arctan (0-1) = 109$$
the base of tangents is $45$, subtract $-45$, add $45$ to $64$

This won't be correct (concatenation of numbers from calculations is not permitted), but this is a way to cheese it, were that allowed

Assuming you can have leading 0's...

$ 2^0 = 1$

$ sqrt81 = 09$

Concatenate the two

$109$

This is also just for fun, using $!$ in a different way than factorial:

Uses $!$ as the binary NOT operator (like in C++ and JavaScript)
$$108space+space!(!2) = 109$$

Explained:

!2 == false, and !false == true. True is numerically represented as $1$. Then you get $108 + 1$, which equals $109$.

We include 3 numbers 1,0,8 to 108, 108 can be written as 107+1
now we have 2 remaining from the list of given numbers we can use as 107+(2*1)=109

Easy, just use a one sided self referencing equation, an ingenious mathematical artefact invented by me just now :

Not correct, because it uses the same digits more than once, but you could get there like this:

$(2 * 8^2) - (8 * 2) - (frac12 * 8) + 1^0 = 109$

Even if it was solved, I couldn't help myself and came up with a boring

$$dfrac218020=109$$