Introduction:
In this problem, we need to find the maximum value of n such that the given expression is perfectly divisible by 21^n.

Solution:

Step 1: Prime Factorization of 21
21 can be factorized as 3 × 7

Step 2: Prime Factorization of the given expression
77 = 7 × 11
42 = 2 × 3 × 7
37 = 37
57 = 3 × 19
30 = 2 × 3 × 5
90 = 2 × 3 × 3 × 5
70 = 2 × 5 × 7
2400 = 2 × 2 × 2 × 2 × 2 × 3 × 5 × 5
2402 = 2 × 7 × 7 × 13
243 = 3 × 3 × 3 × 3
343 = 7 × 7 × 7

Step 3: Re-write the expression in terms of 3 and 7
77 × 3 × 7 × 11 × 2 × 3 × 7 × 37 × 3 × 19 × 2 × 3 × 5 × 2 × 3 × 3 × 5 × 2 × 5 × 7 × 2 × 2 × 2 × 2 × 2 × 3 × 5 × 5 × 2 × 7 × 7 × 13 × 3 × 3 × 3 × 3 × 7 × 7 × 7

Step 4: Count the powers of 3 and 7 in the expression
Powers of 3 = 2 + 1 + 1 + 3 + 1 + 3 + 4 = 15
Powers of 7 = 1 + 1 + 1 + 1 + 1 + 1 + 2 + 2 + 3 + 3 + 3 = 19

Step 5: Determine the maximum value of n
We need to find the maximum value of n such that the given expression is perfectly divisible by 21^n.
Since 21 = 3 × 7, we need to consider the minimum power of 3 and 7 in the expression.
Minimum power of 3 = 15
Minimum power of 7 = 19
Hence, the maximum value of n is 15.

Conclusion:
The maximum value of n such that 77*42*37*57*30*90*70*2400*2402*243*343 is perfectly divisible by 21^n is 15.