It's actually [n!e] - 1 where [x] the greatest integer less than or equal to x and e is the base of the natural logarithm. As nbouscal points out, the total number of heads is H=n!(1+1/1!+ 1/(n-1)!). Using the Taylor series for e^x, H = n!e - n!(1/n! + 1/(n+1)! + ...) = n!e - 1 - r, where r = (1/(n+1) + 1/(n+1)(n+2) ...). The remainder term r is easily seen to be less than 1, and H is and integer, so H=[H+r]=[n!e]-1.

There will be n! of the smallest heads, the ones which don't spawn any new heads. Total heads, though, will be 9 + 9 * 8 + 9 * 8 * 7 + ...

aargh, of course... thanks!

It's actually 9* (1 + 8 * (1 + 7 * (1 + 6 ..... + 1)))

= [9!e] - 1

(because your expression expands to 9 + 9(8) + 9(8)(7) + ... 9! = 9!(1+1/1 + 1/2! + ... 1/8!) = -1 + 9! e - (1/10 + 1/(10 11) + ...))