The given polynomial can be written as the product of several binomials, where each binomial is of the form (x + k.nCk), where k varies from 0 to n.

To find the coefficient of xn in this polynomial, we need to consider all possible combinations of terms that can contribute to the coefficient of xn.

Consider the binomial (x + k.nCk). The term x will contribute to the power of x, while the constant term k.nCk will contribute to the coefficient of xn.

Let's analyze the contribution of each binomial to the coefficient of xn:

1. For the binomial (x + 0.nC0), the constant term is 0.nC0 = 0. This binomial does not contribute to the coefficient of xn.

2. For the binomial (x + 1.nC1), the constant term is 1.nC1 = n. This binomial contributes a term of nxn-1 to the coefficient of xn.

3. For the binomial (x + 2.nC2), the constant term is 2.nC2 = 2n(n-1)/2 = n(n-1). This binomial contributes a term of n(n-1)xn-2 to the coefficient of xn.

4. Similarly, for the binomial (x + 3.nC3), the constant term is 3.nC3 = 3n(n-1)(n-2)/6 = n(n-1)(n-2)/2. This binomial contributes a term of n(n-1)(n-2)/2xn-3 to the coefficient of xn.

5. Continuing this pattern, we can see that for the binomial (x + (2n-1).nCn), the constant term is (2n-1).nCn = (2n-1)n(n-1)(n-2)...(n-n+1)/n! = (2n-1)n!. This binomial contributes a term of (2n-1)n!xn-n to the coefficient of xn.

Now, let's find the coefficient of xn by multiplying all these terms together:

Coefficient of xn = n * n(n-1) * n(n-1)(n-2)/2 * n(n-1)(n-2)(n-3)/3! * ... * (2n-1)n!
= (nC0 * nC1 * nC2 * ... * nCn) * n!

Notice that the terms nC0, nC1, nC2, ..., nCn form the binomial expansion of (1 + 1)n. Therefore, we have:

Coefficient of xn = (1 + 1)n * n!

Simplifying further, we get:

Coefficient of xn = (n + 1)n * n!
= (n + 1)! = (n + 1) * n!

Hence, the correct answer is option 'C', (n + 1)2n.