From: peter.luschny@gmail.com   
      
   This time I am using Sage as the engine.   
      
   Step 1:   
      
   m, n = var('m,n')   
   for k in (0..4):   
       expand(add(m^(k-i)*binomial(n-i-1,k-i) for i in (0..k)))   
      
   [0] 1   
   [1] m*n - m + 1   
   [2] 1/2*m^2*n^2 - 3/2*m^2*n + m^2 + m*n - 2*m + 1   
   [3] 1/6*m^3*n^3 - m^3*n^2 + 11/6*m^3*n + 1/2*m^2*n^2   
       - m^3 - 5/2*m^2*n + 3*m^2 + m*n - 3*m + 1   
      
   Step 2:   
      
   m, n = var('m,n')   
   for k in (0..4):   
       print [k], expand(add(m^i*rising_factorial(n-k,i)/factorial(i) \   
             for i in (0..k)))   
      
   [0] 1   
   [1] m*n - m + 1   
   [2] 1/2*m^2*n^2 - 3/2*m^2*n + m^2 + m*n - 2*m + 1   
   [3] 1/6*m^3*n^3 - m^3*n^2 + 11/6*m^3*n + 1/2*m^2*n^2   
       - m^3 - 5/2*m^2*n + 3*m^2 + m*n - 3*m + 1   
      
   Step 3:   
      
   We conclude, for all m,n,k:   
      
   sum_{i=0..k} m^(k-i)*binomial(n-i-1,k-i) =   
   sum_{i=0..k} m^i*rising_factorial(n-k,i)/i!   
      
   We are interested in the case n=k.   
      
   sum_{i=0..n} m^(n-i)*binomial(n-i-1,n-i) =   
   sum_{i=0..n} m^i*rising_factorial(0,i)/i!   
      
   rising_factorial(0,i) = 1 if i = 0 otherwise 0.   
   Therefore the considered sum has to be 1.   
      
   ====   
      
   So what does Sage return?   
   File "expression.pyx", line 1178,   
   (sage/symbolic/expression.cpp:8057)   
   TypeError: unable to simplify to float approximation   
      
   Also Harald Schilly told me yesterday on Google+:   
      
   "so well, maxima agrees with mathematica   
   var("i n")   
   sum(binomial(n-i-1, n-i), i, 0, n) => 0"   
      
      
   I did not check this.   
      
   CC> Can Alpha display steps for this problem?   
      
   Perhaps it can, but it does not offer this option,   
   at least not for the not-paying user.   
      
   Peter   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)