I suspect that n is declared as unsigned int and you've compiler with a data model that assumes int to be 32 bits wide. The maximum value that can be represented with this type would be 232 - 1 = 4294967295. Anything beyond this value would wrap around. So assigning 4294967296 would become 0, 4294967297 would become 1, and so on.
You have an overflow; since both operands are unsigned int the resulting type would be the same too. The true result of the operation would be 6049417284. Assigning it to an unsigned int would (wrap) and become 1754449988 = 6049417284 - 4294967296. This unsigned int result is assigned to a wider type unsigned long long, which doesn't change the value. It's necessary to understand the difference between the result's type (the type of the expression) and destination type (the type of the variable that is going to hold the result).
Wrap around behaviour (more formally modulo n) in unsigned types is well-defined in C++, so the compiler might not warn you.
Quote from Unsigned Arithmetic:
If an unsigned integer overflows, the result is defined modulo 2w, where w is the number of bits in that particular unsigned integer. By implication, an unsigned integer is never negative.
