ClearAll[a, b];
 a[1] := 2;
 a[2] := a[1] + b;
 b = 2;
 a[n_] := Product[a[i], {i, 1, n - 1}] + b
 list = Table[a[i], {i, 1, 5}]

f = FindSequenceFunction[list]

Table[3^2^n + 1, {n, 0, 3}]

ClearAll[a, b];
 With[{b = 2},
 RecurrenceTable[{a[n] == Product[a[i], {i, 1, n - 1}] + b, a[1] == 2,
 a[2] == a[1] + b}, a, {n, 1, 5}]]

ClearAll[a, b];
 With[{b = 2},
 RSolve[{a[n] == Product[a[i], {i, 1, n - 1}] + b, a[1] == 2,
   a[2] == a[1] + b}, a[n], n]]

ClearAll[a, b];
 prod[n_] := Module[{a},
  a[1] := 2; b := 2; a[2] := a[1] + b; k := (n - 1);
  a[i_] := Product[a[j], {j, 1, k}] + b;
  a[n]
  ]

prod[1]

prod[4]