Weapons of Math Instruction

Javascript ES7 49 bytes

a=>a.reduce((c,d,e)=>[c**d,c+d,c-d,c*d,c/d][e%5])

Saved 9 bytes thanks to Dom Hastings, saved another 6 thanks to Leaky Nun

Uses the new exponentiation operator.

Haskell, 76 65 64 62 bytes

Thanks to @Damien for removing another two bytes=)

f(u:v)=foldl(\x(f,y)->f x y)u(zip(v>>[(+),(-),(*),(/),(**)])v)

This uses the >> which here just appends the list [(+),...] to itself length v times. The rest still works still the same as the old versions.

Old versions:

These solutions make use of the infinite lists, as cycle[...] just repeats the given list infinitely. Then it basically gets ziped with the list of numbers, and we just fold (reduce in other languages) the zipped list via a lambda, that applies the operators to the accumulator/current list element.

f(u:v)=foldl(\x(f,y)->f x y)u(zip(cycle[(+),(-),(*),(/),(**)])v)

f(u:v)=foldl(\x(y,f)->f x y)u(zip v(cycle[(+),(-),(*),(/),(**)]))

f l=foldl(\x(y,f)->f x y)(head l)(zip(drop 1l)(cycle[(+),(-),(*),(/),(**)]))

Pyke, 22 21 bytes

lt5L%"+-*/^"L@\RJQ_XE

Try it here!

lt5L%                 -    map(len(input)-1, %5)
     "+-*/^"L@        -   map(^, "+-*/^"[<])
              \RJ     -  "R".join(^)
                    E - pyke_eval(^, V)
                 Q_X  -  splat(reversed(input))