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Question

Prove that if $$n$$ and $$r$$ are positive integers
$${ n }^{ r }-n{ \left( n-1 \right)  }^{ r }+\dfrac { n\left( n-1 \right)  }{ 2! } { \left( n-2 \right)  }^{ r }-\dfrac { n\left( n-1 \right) \left( n-2 \right)  }{ 3! } { \left( n-3 \right)  }^{ r }+\cdots $$
is equal to $$0$$ if $$r$$ be less than $$n$$, and to $$n!$$ if $$r=n$$.


Solution

We have 
$${ \left( { e }^{ x }-1 \right)  }^{ n }={ \left( x+\dfrac { { x }^{ 2 } }{ 2! } +\dfrac { { x }^{ 3 } }{ 3! } +\dfrac { { x }^{ 4 } }{ 4! } +\cdots  \right)  }^{ n }$$
$$={ x }^{ n }+$$ terms containing higher powers of $$x$$ ..... $$(1)$$
Again, by the Binomial Theorem,
$${ \left( { e }^{ x }-1 \right)  }^{ n }={ e }^{ nx }-n{ e }^{ \left( n-1 \right) x }+\dfrac { n\left( n-1 \right)  }{ 1\cdot 2 } { e }^{ \left( n-2 \right) x }-\cdots $$         ......(2).
By expanding each of the terms $${ e }^{ nx },{ e }^{ \left( n-1 \right) x },\dots $$ we find that the coefficient of $${ x }^{ r }$$ in (2) is
$$\dfrac { { n }^{ r } }{ r! } -n\cdot \dfrac { { \left( n-1 \right)  }^{ r } }{ r! } +\dfrac { n\left( n-1 \right)  }{ 2! } \cdot \dfrac { { \left( n-2 \right)  }^{ r } }{ r! } -\dfrac { n\left( n-1 \right) \left( n-2 \right)  }{ 3! } \cdot \dfrac { { \left( n-3 \right)  }^{ r } }{ r! } +\cdots $$
and by equating the coefficients of $${ x }^{ r }$$ in (1) and (2) the result follows.

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