Optimize pow when self is a power of two

if base == 2 ** k, then
   (2 ** k) ** n
== 2 ** (k * n)
== 1 << (k * n)

Author: Kmeakin

library/core/src/num/uint_macros.rs

@@ -3318,6 +3318,16 @@ macro_rules! uint_impl {
             let mut acc = 1;
 
             if intrinsics::is_val_statically_known(exp) {
+                // change of base:
+                // if base == 2 ** k, then
+                //    (2 ** k) ** n
+                // == 2 ** (k * n)
+                // == 1 << (k * n)
+                if self.is_power_of_two() {
+                    let k = self.ilog2();
+                    return 1 << (k * exp)
+                }
+
                 while exp > 1 {
                     if (exp & 1) == 1 {
                         acc = acc * base;