How to prove the equality $\sum_{j=0}^n (x)^j (-1)^{n-j} \left\{{n \atop j}\right\} = x^n$? How do you prove $$\sum_{j=0}^n (x)^j (-1)^{n-j} \left\{{n \atop j}\right\} = x^n,$$
where $(x)^j=x(x+1)...x(x+j-1)$ and $\left\{{n \atop j}\right\}$ is a Stirling number of the second kind?

There's a simple but not very well-known formula on row sums for number triangles that can be applied here.  I believe it is due to Neuwirth.  Suppose you have a triangle of numbers $R(n,k)$ with $R(0,0) = 1$ and, for $n \geq 1$, satisfying $$R(n,k) = (\alpha (n-1) + \beta k + \gamma) R(n-1,k) + (\alpha' (n-1) + \beta' (k-1) + \gamma') R(n-1,k-1).$$
There are several interesting number triangles that are special cases, such as the binomial coefficients and both kinds of Stirling numbers.  
Anyway, the formula is that if $\beta + \beta' = 0$ then $$\sum_{k=0}^n R(n,k) = \prod_{i=0}^{n-1} \left((\alpha + \alpha')i + \gamma + \gamma'\right).$$
It is easy to verify that $R(n,k) = \left\{ { n \atop k } \right\} x^{\underline{k}}$ satisfies the above recurrence with $\beta = 1, \beta' = -1, \gamma' = x$ and all other parameters $0$.  Thus the row sum formula yields Derek's reformulation of the problem (Eq. (1)):
$$\sum_{k=0}^n \left\{ { n \atop k } \right\} x^{\underline{k}} = \prod_{i=0}^{n-1} x = x^n.$$
