Multiplying elements of a list

FoldList[Times, l] (* or *)
FoldList[Times]@l

{2, 6, 24, 120, 720}

Also:

Exp @ Accumulate @ Log @ l

{2, 6, 24, 120, 720}

list = Range[2, 6]

(* {2, 3, 4, 5, 6} *)

To keep the factors separate, use Inactive with kglr's solution

list2 = Rest@FoldList[Inactive@Times, list[[1]], Rest@list]

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Activate produces the result

list3 = list2 // Activate

(* {6, 24, 120, 720} *)

Or use NonCommutativeMultiply to hold the factors

list4 = Rest@FoldList[NonCommutativeMultiply, list[[1]], Rest@list]

(* {2 ** 3, 2 ** 3 ** 4, 2 ** 3 ** 4 ** 5, 2 ** 3 ** 4 ** 5 ** 6} *)

Then Apply Times to get the final result

list5 = Times @@@ list4

(* {6, 24, 120, 720} *)

For this specific case (sequential numbers), the result is just Factorial

list3 == list5 == Factorial /@ Rest@list

(* True *)

Use Part to access any element of the result, e.g.,

list3[[1]]

(* 6 *)

This can also be solved using recursion. The function cumProd (cumulative product) can be defined as:

list = Range[10];
cumProd[n_] := cumProd[n - 1]*list[[n]];
cumProd[1] = list[[1]];

To use:

cumProd[6]
720

gives the 6th "level" of the product. Of course, list can be any set of numbers. Applying this to the whole list:

cumProd/@Range[Length[list]]
{1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800}

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