My idea was:

$1-P(\text{No element is assigned to its correct spot})=1-\frac{(n-1)!}{n!}=\frac{n-1}{n}$

Numerator: For the first Element there are n-1 incorrect spots, for the second there are n-2 incorrect spots, ... . So alltogether (n-1)! possible incorrect assignments

Denominator: I can assign the first element to n spots, the second to (n-1) spots, ... . So alltogether n! possible assignments

Apparently this solution is wrong. I understand correct solution ($1-\frac{1}{e}$ as n tends to infinity) and its derivation but I don't understand why my idea leads to an incorrect answer.

Thanks for your help!