Given that p≥ -1 , prove by induction that, for all integers n≥1 , (1+p)^k ≥ 1+k*p.

First of all, we need to show that the statement is true for the base case n=1. For this case the expression becomes: 1+p≥1+p, which is clearly true as both sides are equal and hence solve the inequality.  Next for a proof by induction, we want to assume that the statment is true for n=k, and hence show it is true for n=k+1. So we start off with the statement: (1+p)^k ≥ 1+kp Multiplying both sides by (1+p) we obtain: (1+p)^k+1≥ (1+kp)(1+p)        () Remember though we have to be careful when multiplying through an inequality - if we multiply through by a negative number the inequality sign flips round. However the question states that p≥-1, which means (1+p) is positive so the above statement is true unless p=-1. We should look more carefully at the case p=-1, as multiplying both sides by zero is an 'illegal move' - it just makes both sides zero. When p=-1 the orginial expression is 0≥1-n, as n≥1 it is easy to see that the RHS is at most 0 when n=1 and so the inequality holds for all n, when p=-1.  Returning to equation () and multiplying out the RHS we get: (1+p)^k+1≥ 1+p+kp + kp² = 1+(1+k)p + kp²  ≥ 1 + (1+k)p The last inequality is true because k is a positive integer and p^2≥0 so kp²≥0.  So we have shown that (1+p)^k+1≥1 + (1+k)*p which is the statement for n=k+1.  To finish the proof by induction all we need to do is state, 'We have shown that if the statement is true for n=k then it is true for n=k+1. As it is true for n=1, it is therefore true for all positive integers.'