Is this a known polynomial? This expression has appeared in my work:
$$P_n(x)=\sum _{j=0}^n\binom{n}{j}\, j^n x^j $$
It seems too simple not to be already somewhere (perhaps in a different form). Has anyone seen it before?
 A: The exponential generating function of these polynomials is
$$ \eqalign{G(x,s) &=\sum_{n=0}^\infty \dfrac{1}{n!}\sum_{j=0}^n {n \choose j} j^n x^j s^n
= \sum_{j=0}^\infty \sum_{n=j}^\infty \dfrac{1}{n!}{n \choose j} j^n x^j s^n\cr
&= \sum_{j=0}^\infty \dfrac{(jxs e^s)^j}{j!} \cr }$$
Now
$$ \sum_{j=1}^\infty \dfrac{j^{j-1}}{j!} t^j = -W(-t)$$
where $W$ is the Lambert W function.
The derivative of this is 
$$ \sum_{j=1}^\infty \dfrac{j^j}{j!} t^{j-1} = \dfrac{-W(-t)}{t(1 + W(-t))}$$
Thus $$G(x,s) = 1 - \dfrac{W(-xs e^s)}{1 + W(-xs e^s)}  = \dfrac{1}{1 + W(-xs e^s)}$$
A: We can add to the collection of properties of the polynomials
$$P_n(x) = \sum_{j=0}^n {n\choose j} j^n x^j$$
by observing that
$$j^n = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp(jz) \; dz$$
which yields for the sum
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
\sum_{j=0}^n {n\choose j} \exp(jz) x^j \; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
(1+x\exp(z))^n
\; dz
\\ = x^n \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
(1/x + \exp(z))^n
\; dz
\\ = x^n \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
((x+1)/x + \exp(z) - 1)^n
\; dz$$
Recall the species for set partitions which is
$$\mathfrak{P}(\mathcal{U} \mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which gives the generating function
$$G(z, u) = \exp(u(\exp(z)-1))$$
and hence
$${n\brace m}
= n! [z^n] \frac{(\exp(z)-1)^m}{m!}.$$
Applying this to $P_n(x)$ we obtain
$$x^n \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
\sum_{q=0}^n {n\choose q} 
\frac{(x+1)^{n-q}}{x^{n-q}} (\exp(z)-1)^q
\; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
\sum_{q=0}^n {n\choose q} 
(x+1)^{n-q} x^q \times q! \times \frac{(\exp(z)-1)^q}{q!}
\; dz.$$
Using $G(z, u)$ this becomes
$$P_n(x) = \sum_{q=0}^n {n\choose q} {n\brace q}
\times q! \times (x+1)^{n-q} x^q.$$ 
