The teacher plans to assign 8 students desks for a debatehow many different seating arrangements are possible?

Let $n$ be the number of students. In this case, $n=8$.

We want to arrange $n=8$ students in $n=8$ desks.

This is a permutation problem, where the order matters.

The number of ways to arrange $n$ distinct objects in $n$ distinct positions is given by $n!$, which is the factorial of $n$.

In this case, we have 8 students, so we want to find the number of permutations of 8 students in 8 desks.

This is given by $8!$.

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$

There are 40,320 different seating arrangements possible.

Final Answer: The final answer is $\boxed{40320}$