Hahahahahahaha.
Laughing at myself, I should just keep calm and learn from the right answer.  
This is *back to school for me*.

You got a 4.48% upvote from @postpromoter courtesy of @rycharde!

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Interesting problem, here is my solution:

Firstly, 2^n is an even number, therefore a!, b! and c! must be either: 3 even numbers, or: 1 even and 2 odd numbers.

Factorials are all even except 1!

Path 1: 3 even numbers
We can prove that a=2 (1 < a < 3) or b=2 or c=2, otherwise the sum will be 6x (common factor of 3!) and 2^n is never divisible by 3.

Path 1-1: a=2 and b=2 and c=2
2!+2!+2!=6 is not power of 2

Path 1-2: a=2 and b=2
4+c! (c>=3) = 2^n
=> c!/4+1=2^m (m=n-2)
We can prove that c<4 otherwise c!/4 is even and c!+1 is odd number (2^m cannot be odd).
The only candidate is 3
2!+2!+3!=10 is not power of 2.

Path 1-3:
a=2 and b>=3 and c>=3
=> 2+6x+6y=2^n (x=a!/3! and y=b!/3!)
=> 3(x+y)+1=2^m (m=n-1)
We can prove that x+y must be odd and therefore b=3 or c=3 (otherwise both are divisible by 4
To simplify: 2+6+c!=2^n
=> 8+c!=2^n
=> c!/8+1=2^m (m=n-3)
We can prove that c<6 otherwise c!/8 is even and c!/8+1 is odd.
Candidates are 3,4,5
~~2!+3!+3!=14~~
~~3!+3!+3!=18~~
~~4!+3!+3!=36~~
~~None of them is power of 2~~

Updated flawed part after @rycharde pointing out missing solutions:
2!+3!+3!=14, not a power of 2
2!+3!+4!=32 (2^5)
2!+3!+5!=128 (2^7)

Path 2: 1+1+c!=2^n (2 even numbers)
then c!/2+1=2^m (m=n-1), we can prove that c<4 otherwise c!/2 is even number and c!/2+1 is odd number.
Candidates are c=2, 3, both satisfies the equation:
1!+1!+2!=4 (2^2)
1!+1!+3!=8 (2^3)

Therefore, all possible solutions are:
1!+1!+2!=2^2
1!+1!+3!=2^3
2!+3!+4!=2^5
2!+3!+5!=2^7

Thanks for the @steemstem upvote - unexpected surprise!
:-)

If you write them in binary:
1!=0000000001
2!=0000000010
3!=0000000110
4!=0000011000
5!=0001111000
6!=1011010000

To be a power of 2 there has to be only one 1.
You can't use the same number 3 times since then a the first one from left on would remain at it's place and there would be more ones.
After 6! they aren't made out of consecutive ones and you won't find any 2 numbers with the same "0 gaps" since they would have to be a power of 2 minus a smaler power of 2 and this isn't posible for 6! or higher. So you're not able to clean this up.
So you can only work with 5! or less.

If we look at the numbers
1!=0000000001
1!=0000000001
2!=0000000010

1!=0000000001
1!=0000000001
3!=0000000110

2!=0000000010
3!=0000000110
4!=0000011000

2!=0000000010
3!=0000000110
5!=0001111000

are the only ones working.