[civild]

You gave it away with a slight error - the sequence is 0, 1, 2, 720...

For the curious, the sequence follows factorial (!) applied twice, i.e. (0!)!, (1!)! etc. (3!)! is 720, and (4!)! is 6.20448402 × 10^23.

[benjoffe]

In that case the sequence should be 1, 1, 2, 720... as zero factorial equals one.

https://www.google.com/search?q=0!

[civild]

Oops, quite right!

[dedward]

0 is correct pragmatacally, and makes the puzzle harder for those not wise that 0!=1 by the definition of factorial. (as opposed to proof.... factorial being a shorthand for math and ths being convenient. if there were a proof it would be a theorem, which its not.)

i knew factorial but i had to read up on 0!, news to me too.

on another note, without context, we could say there is an infinite set of functions that satisfy any such question.

"what could come next and why" or something.

edit: 1 is indeed correct for the sequence, ignore that part... my bad.

[waqf]

No, the sequence is correct as originally posted and the rule is:

0, 1!, (2!)!, ((3!)!)!, (((4!)!)!)!, ...

[reitzensteinm]

While the zero is off in the GP's post, he technically could have been talking about the series of n!!!

[anonymoushn]

I have (0!)! = (1!)! = 1 over here.

[mikeash]

His sequence (corrected to 1, 1, 2, 720!, ...) would work for the factorial applied three times. Although the ! pretty much gives it away then.