Create Sum of all factorials of n!

The sum of all factorials of n!

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+Following are the steps for an efficient solution:
+
+n! can be represented as :-  n! = (a1^p1) (a2^p2)x...(ak^pk).
+
+where ak is the prime divisor lesser than n and pk is the highest power which can divide n!.
+
+Through sieve find the prime number and the highest power can be easily find out by :
+
+countofpower =  [n/a] + [n/a^2] + [n/a^3] +...... or   `  while (n)
+{
+  n/ = a;
+  ans += n
+}
+Count Of Factors = (ans1 +1)*(ans2 +1)*....(ansk +1)
+
+After calculating this last step is the sum :
+
+SUM = product of all (pow(ak,pk+1)-1)/(ak-1);
+ex = 4! 
+4! = 2^3 * 3^1;
+count of factors = (3+1)*(1+1) = 8 (1,2,3,4,6,8,12,24)
+sum = ( 1 + 2 + 4 + 8)*(1 + 3) = 60.