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They can all be expressed as themselves to the power one, but to make the problem interesting let's specify that the exponent has to be an integer greater than 2. The highest exponent we are going to need is the highest integer solution to 2p < 1000. Which is in fact p = 9. [29 = 512.] The highest base we are going to need is the highest integer solution to b2 < 1000. This is given by b = 31. [312 = 961.] But the only real way to answer the question is just to get stuck in. 1 = 12

4 = 22

8 = 23

9 = 32

16 = 24 = 42

25 = 52

32 = 25

36 = 62

49 = 72

64 = 26 = 43 = 82

81 = 34 = 92

100 = 102

121 = 112

125 = 53

128 = 27

144 = 122

169 = 132

196 = 142

216 = 63

225 = 152

243 = 35

256 = 28 = 44 = 162

289 = 172

324 = 182

343 = 73

361 = 192

400 = 202

441 = 212

484 = 222

512 = 29 = 83

529 = 232

576 = 242

625 = 54 = 252

676 = 262

729 = 36 = 93 = 272

784 = 282

841 = 292

900 = 302

961 = 312

1000 = 103

Well, they're all there. 41 different numbers in total.

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