Question

how many digits would be in the product 2*2*2........*2(30times)

Answer 1

2² x 2³ = 2^5 = 32 

(2^5)² = 2^10 = 32² = 1024 = 1.024 x 10³ 

(2^10)³ = 2^30 = (1.024x 10³)³ = 1.024³ x 10^9 

1.024³ generates a number with just one digit in front of the decimal point. 
Such a number when multiplied by 10^9 will have 10 digits in front of the decimal point and all the fractional decimals will have been cleared.