ABE is the only candidate key.
Start with the tests for 1NF, and show yourself that the existing relation R satisfies 1NF. When you find a normal form that R does not satisfy, fix it.
For example, in testing for 2NF, you see that the FD AB->C is a partial key dependency. (ABE is the only candidate key; C is dependent on only part of that key.) Use projection to remove C from R.
- R { ABE CD} is what we started with. After removing C by projection we have
- R1 { ABE D}, and
- R2 { AB C}
Repeat until all your tables are in 5NF. (R1 is not in 2NF; R2 is in 5NF.)