Why must x be positive in the proof that x^n = ∑(x!/(x!-k!))S(n,k)?

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So the question was, Let x > 0.

Prove that x[itex]^{n}[/itex] = [itex]\sum[/itex] [itex]\frac{x!}{x!-k!}[/itex] S(n, k).

Where the sum goes from k = 1 to n and S(n, k) is th Stirling numbers.

I believe I have proven what I needed to, but my question is why does x have to be greater than 0? Couldn't we define a function that maps [n] to {-x, ..., 1}.
 
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Punkyc7 said:
So the question was, Let x > 0.

Prove that x[itex]^{n}[/itex] = [itex]\sum[/itex] [itex]\frac{x!}{x-k!}[/itex] S(n, k).

Where the sum goes from k = 1 to n and S(n, k) is th Stirling numbers.

I believe I have proven what I needed to, but my question is why does x have to be greater than 0? Couldn't we define a function that maps [n] to {-x, ..., 1}.

Is this equation accurate? The kth term blows up when x = k! for some k in {1,2,...,n}.

RGV