Problem in Finding Number of Trailing Zeroes

22! will have 4 trailing 0's and = 1124000727777607680000

In scientific notation, this would be = 1.12400072777760768  x 10^21

And 21!  = 51090942171709440000

So....putting this together, we have

[ 1.12400072777760768 x 10^21] ^ 51090942171709440000  =

[ 1.12400072777760768]^ 51090942171709440000  x [10^21]^51090942171709440000

The first exponentiation provides no extra trailing 0's.....the second one provides.....

21 x 51090942171709440000  = 1072909785605898240000   trailing zeroes  (1)

Likewise  27! provides 6 trailing zeroes and is = 10888869450418352160768000000

Again, in scientific notation, this would written as 1.0888869450418352160768 x 10^28

And 26!  = 403291461126605635584000000

So...putting the second part all together, we have....

[1.0888869450418352160768 x 10^28]^403291461126605635584000000 =

[1.0888869450418352160768] ^ 403291461126605635584000000 x [10^28]^403291461126605635584000000

Again, the first part provides no extra trailing 0's, whereas the second part provides

28 * 403291461126605635584000000  = 11292160911544957796352000000  trailing 0's  (2)

Note that....if for instance, we multiplied 10,000 x 10,000.....the number of trailing 0's in the result is just the sum of the trailing 0's in both of the multiped numbers

So adding (1) and (2)   together, we have........

1072909785605898240000 + 11292160911544957796352000000 =

11,292,161,984,454,743,402,250,240,000   trailing zeroes   !!!!

BTW....this number is pronounced as :

11 octillion 292 septillion 161 sextillion 984 quintillion 454 quadrillion 743 trillion 402 billion 250 million 240 thousand.....[and that's a mouthful !!!]

I came up with: 10^540,000 zeros!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!.By the way, the answer to your expression is =10^(10^28.05335034318211)