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?

To remove the $(-1)^{n-j}$ term we replace $x$ by $-x$ and prove
$$x^n = \sum_{j=0}^n \left\{ { n \atop j } \right\} (x)_j , \quad (1)$$
where $(x)_j = x(x-1)(x-2) \cdots (x-j+1)$ and $(x)_0 = 1.$
We proceed by induction. Note that $(1)$ is true for $n=0$ and so we take $n>0$ and assume $(1)$ is true for $n-1.$
$$\begin{align*}
x^n &= x \cdot x^{n-1} \\
&= x \sum_{j=0}^{n-1} \left\{ { n-1 \atop j } \right\} (x)_j \\
&= \sum_{j=0}^{n-1} \left\{ { n-1 \atop j } \right\} (x)_j (x-j + j) \\
&= \sum_{j=0}^{n-1} \left\{ { n-1 \atop j } \right\} (x)_{j+1} +
\sum_{j=0}^{n-1}  j \left\{ { n-1 \atop j } \right\} (x)_j \\
&= \sum_{j=1}^n \left\{ { n-1 \atop j-1 } \right\} (x)_j +
\sum_{j=0}^{n-1}  j \left\{ { n-1 \atop j } \right\} (x)_j \\
&= \sum_{j=0}^n \left\{ { n \atop j } \right\} (x)_j.
\end{align*}$$
The last equality follows from the recurrence relation for the Stirling numbers of the second kind:
$$ \left\{ { n \atop j } \right\} = \left\{ { n-1 \atop j-1 } \right\} +
j \left\{ { n-1 \atop j } \right\}.$$ 
