(expi)^i

can someone tell me how can we calculate (expi)^i.
there is one thought to say that it is equal to exp(-1) but this is not true.
i think that one has to use the euler's equation : exp(ix)=cosx+isinx. but in this case, x=1? and how can we calculate after (cos1+isin1)^i?
anyway that was my way of thinking, of course someone specialised on maths knows better than me? Any ideas?

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asked by natassa

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