Showing equivalence of two binomial expressions I wish to show that $\sum_{k=0}^n {n\choose k}(\alpha + k)^k (\beta + n - k)^{(n-k)} = \sum_{k=0}^n {n\choose k}(\gamma + k)^k (\delta + n - k)^{(n-k)}$ given that $\alpha + \beta = \gamma + \delta$.
Both sides appear to be the n-th coefficient in the product of some exponential generating series, namely $F(y,x) = \sum_{n=0}^{\infty} (y+n)^n / n!$, and that $F(y,x) = \exp(y\times T(x))$ for some series $T(x)$. It seems to be exponential because we could then have $F(a+b,x) = F(a,x) F(b,x)$.
I have thought to use the Lagrange inversion theorem in a manner similar to some proofs of Abel's binomial theorem. Namely, if $T(x)$ is the solution to $T(x) = x \phi (T(x))$ for some invertible $\phi(x)$ in $\mathbb{Z} [[x]]$, then for $f(y,x) = exp(yx)$ we would have:
$[x^n]f(yT(x)) = \frac{1}{n} [x^{n-1}] f'(x) \phi(x)^n = (y+n)^n / n!$.
The issue is then in showing such a $\phi(x)$ exists. Unfortunately, my attempts using this approach have not been successful. Could I have some hints or suggestions on how to proceed?
Regards,
Garnet
 A: We  can  prove this  using  a  close relative of the  labelled  tree
function that is known from combinatorics.
This  will  provide a  closed form  of  the exponential
generating function of the four terms that are involved.
The species of labelled trees has the specification
$$\mathcal{T} = 
\mathcal{Z} \times \mathfrak{P}(\mathcal{T})$$
which gives the functional equation
$$T(z) = z \exp T(z).$$
Now consider the function
$$Q_\rho(z) = \frac{\exp(\rho T(z))}{1-T(z)}$$
with $\rho$ a real parameter.
Extracting coefficients via Lagrange inversion we have
$$Q_n
= n! \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
\frac{\exp(\rho T(z))}{1-T(z)} dz.$$
Put $T(z)=w$ so that $z=w/\exp(w) = w\exp(-w)$ and 
$dz = \exp(-w) - w\exp(-w)$ 
to get
$$n! \frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{\exp(w(n+1))}{w^{n+1}} 
\frac{\exp(\rho w)}{1-w} (\exp(-w) - w\exp(-w)) dw
\\ = n! \frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{\exp(wn)\exp(w\rho)}{w^{n+1}} 
\frac{1}{1-w} (1 - w) dw
\\ = n! \frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{\exp(w(\rho+n))}{w^{n+1}} dw.$$
But we have
$$n! [w^n] \exp(w(\rho+n)) = 
n! \times \frac{(\rho+n)^n}{n!} = (\rho+n)^n$$
which means that $Q_\rho(z)$ is the exponential generating function
of $(\rho+n)^n.$
The equality that we seek to  prove is a convolution of two exponential
generating functions on the left  and on the right and to verify it we 
must show that
$$Q_\alpha(z) Q_\beta(z) = Q_\gamma(z) Q_\delta(z).$$
But this is simply
$$\frac{\exp(\alpha T(z))}{1-T(z)}
\frac{\exp(\beta T(z))}{1-T(z)} =
\frac{\exp(\gamma T(z))}{1-T(z)}
\frac{\exp(\delta T(z))}{1-T(z)}$$
which is
$$\frac{\exp((\alpha+\beta) T(z))}{(1-T(z))^2} =
\frac{\exp((\gamma+\delta) T(z))}{(1-T(z))^2}$$
which holds since $\alpha+\beta = \gamma+\delta.$
The labelled tree function recently appeared at this MSE link.
