Update sol1.py

Author: MaximSmolskiy

Diff for: project_euler/problem_122/sol1.py

@@ -8,20 +8,21 @@
                                                n x n x ... x n = n^15.
 
 But using a "binary" method you can compute it in six multiplications:
+
                                                          n x n = n^2
                                                      n^2 x n^2 = n^4
                                                      n^4 x n^4 = n^8
                                                      n^8 x n^4 = n^12
                                                     n^12 x n^2 = n^14
                                                       n^14 x n = n^15
 
-<However it is yet possible to compute it in only five multiplications:
+However it is yet possible to compute it in only five multiplications:
 
-n x n = n^2
-n^2 x n = n^3
-n^3 x n^3 = n^6
-n^6 x n^6 = n^{12}
-n^12 x n^3 = n^{15}
+                                                                n x n = n^2
+                                                              n^2 x n = n^3
+                                                            n^3 x n^3 = n^6
+                                                            n^6 x n^6 = n^12
+                                                           n^12 x n^3 = n^15
 
 We shall define m(k) to be the minimum number of multiplications to compute n^k; for example m(15) = 5.