Tetration, represented as \${}^ba\$, is repeated exponentiation. For example, \${}^32\$ is \$2^{2^2}\$, which is \$16\$.
Given two numbers \$a\$ and \$b\$, print \${}^ba\$.
1 2 -> 1 2 2 -> 4 5 2 -> 3125 3 3 -> 7625597484987 etc. Scientific notation is acceptable.
Remember, this is code-golf, so the code with the smallest number of bytes wins.
Prompt A,B A For(C,2,B A^Ans End 
double c(int a,int b){return b>0?Math.pow(a,c(a,b-1)):1;} Ungolfed & test code:
class M{ static double c(int a, int b){ return b > 0 ? Math.pow(a, c(a, b-1)) :1; } public static void main(String[] a){ System.out.println(c(1, 2)); System.out.println(c(2, 2)); System.out.println(c(5, 2)); System.out.println(c(3, 3)); } } Output:
1.0 4.0 3125.0 7.625597484987E12 
double t(int x,int n){return n?pow(x,t(x,n-1)):1;} Straightforward from the definition of Tetration.
(within the bounds of bash integer data type)
Golfed
E() { echo $(($(printf "$1**%.0s" `seq 1 $2`)1));} Explanation
Build expression with printf, e.g. E 2 5:
2**2**2**2**2**1 then use bash built-in arithmetic expansion to compute the result
Test
E 1 2 1 E 2 2 4 E 5 2 3125 E 3 3 7625597484987 
filter p ($a){[math]::Pow($a,$_)};iex (,$args[0]*$args[1]-join"|p ") This is the shortest of the three approaches I tried, not that great overall though, i'm 100% sure there's a shorter approach but the few things I tried somehow ended up with slightly more bytes.
PS C:\++\golf> (1,2),(2,2),(5,2),(3,3) | % {.\sqsq $_[0] $_[1]} 1 4 3125 7625597484987 Sadly Powershell has no built-in ^ or ** operator, or it would be a clean 32/33 byte answer, i.e.
iex (,$args[0]*$args[1]-join"^")
Axiom 70 bytes
l(a,b)==(local i;i:=1;r:=a;repeat(if i>=b then break;r:=a^r;i:=i+1);r) this less golfed
l(a,b)== local i i:=1;r:=a;repeat(if i>=b then break;r:=a^r;i:=i+1) r (3) -> [l(1,2),l(2,2),l(5,2),l(3,3),l(4,3)] (3) [1, 4, 3125, 7625597484987, 13407807929942597099574024998205846127479365820592393377723561443721764030_ 0735469768018742981669034276900318581864860508537538828119465699464336490_ 06084096 ] Type: List PositiveInteger local i; could be removed. \$\endgroup\$ (fn[a b](last(take a(iterate #(apply *(repeat % b))b)))) Maybe there is a shorter way via apply comp?
{[**] $^a xx$^b} $^a and $^b are the arguments to this anonymous function. $^a xx $^b is a list of the first argument replicated a number of times given by the second argument. [**] reduces that list with the exponentiation operator. The reduction respects the right-associativity of the operator.
3 3 -> 7625597484987\$\endgroup\$3^3^3automatically means3^(3^(3)). See en.wikipedia.org/wiki/Order_of_operations, where it says "Stacked exponents are applied from the top down, i.e., from right to left." \$\endgroup\$