Simple Task Solved Thrice

J, 16 bytes

Function 1, 5 bytes

p:^i.

Function 2, 6 bytes

+p:^i.

Function 3, 5 bytes

>:^i.

How it works

Function 1

p:^i.     Right argument: y

   i.     Compute (0 ... y-1).
p:        Compute P, the prime at index y (zero-indexed).
  ^       Return all powers P^e, where e belongs to (0 ... y-1).

Since P is prime and P > y, y cannot divide P^e.

Function 2

+p:^i.    Right argument: y

 p:^i.    As before.
+         Add y to all results.

If y divided P^e + y, it would also divide P^e + y - y = P^e.

Function 3

>:^i.     Right argument: y

   i.     Compute (0 ... y-1).
>:        Compute y+1.
  ^       Return all powers (y+1)^e, where e belongs to (0 ... y-1).

If y divided (y+1)^e some prime factor Q of y would have to divide (y+1)^e.

But then, Q would divide both y and y+1 and, therefore, y + 1 - y = 1.

Pyth, 17 16 bytes

5 bytes:

^LhQQ

Outputs:

2: [1, 3]
3: [1, 4, 16]
4: [1, 5, 25, 125]

6 bytes:

mh*QdQ

Outputs:

2: [1, 3]
3: [1, 4, 7]
4: [1, 5, 9, 13]

5 bytes:

|RhQQ

Outputs:

2: [3, 1]
3: [4, 1, 2]
4: [5, 1, 2, 3]

Alternate version, in increasing order: -ShQQ

Dyalog APL, 16 17 bytes

1+⊢×⍳

⊢+1+⊢×⍳

1+⊢*⍳