How can one proof the equality
\sum\limits_{v=0}^k \frac{k^v}{v!}=\sum\limits_{v=0}^k \frac{v^v (k-v)^{k-v}}{v!(k-v)!}
for k\in\mathbb{N}_0?
Induction and generating functions don't seem to be useful.
The generation function of the right sum is simply f^2(x) with \displaystyle f(x):=\sum\limits_{k=0}^\infty \frac{(xk)^k}{k!}
but for the left sum I still don't know.
It is \displaystyle f(x)=\frac{1}{1-\ln g(x)} with \ln g(x)=xg(x) for \displaystyle |x|<\frac{1}{e}.
Recall the combinatorial class of labeled trees which is
\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}}\mathcal{T} = \mathcal{Z}\times \textsc{SET}(\mathcal{T})
which immediately produces the functional equation
T(z) = z \exp T(z) \quad\text{or}\quad z = T(z) \exp(-T(z)).
By Cayley's theorem we have
T(z) = \sum_{q\ge 1} q^{q-1} \frac{z^q}{q!}.
This yields
T'(z) = \sum_{q\ge 1} q^{q-1} \frac{z^{q-1}}{(q-1)!} = \frac{1}{z} \sum_{q\ge 1} q^{q-1} \frac{z^{q}}{(q-1)!} = \frac{1}{z} \sum_{q\ge 1} q^{q} \frac{z^{q}}{q!}.
The functional equation yields
T'(z) = \exp T(z) + z \exp T(z) T'(z) = \frac{1}{z} T(z) + T(z) T'(z)
which in turn yields
T'(z) = \frac{1}{z} \frac{T(z)}{1-T(z)}
so that
\sum_{q\ge 1} q^{q} \frac{z^{q}}{q!} = \frac{T(z)}{1-T(z)}.
Now we are trying to show that
\sum_{v=0}^k \frac{v^v (k-v)^{k-v}}{v! (k-v)!} = \sum_{v=0}^k \frac{k^v}{v!}.
Multiply by k! to get
\sum_{v=0}^k {k\choose v} v^v (k-v)^{k-v} = k! \sum_{v=0}^k \frac{k^v}{v!}.
Start by evaluating the LHS.
Observe that when we multiply two
exponential generating functions of the sequences \{a_n\} and
\{b_n\} we get that
A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\ = \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!} = \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}
i.e. the product of the two generating functions is the generating
function of \sum_{k=0}^n {n\choose k} a_k b_{n-k}.
In the present case we have
A(z) = B(z) = 1 + \frac{T(z)}{1-T(z)} = \frac{1}{1-T(z)} by inspection.
We added the constant term to account for the fact that v^v=1 when
v=0 in the convolution. We thus have
\sum_{v=0}^k {k\choose v} v^v (k-v)^{k-v} = k! [z^k] \frac{1}{(1-T(z))^2}.
To compute this introduce
\frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+1}} \frac{1}{(1-T(z))^2} \; dz
Using the functional equation we put z=w\exp(-w) so that dz = (\exp(-w)-w\exp(-w)) \; dw and obtain
\frac{k!}{2\pi i} \int_{|w|=\gamma} \frac{\exp((k+1)w)}{w^{k+1}} \frac{1}{(1-w)^2} (\exp(-w)-w\exp(-w)) \; dw \\ = \frac{k!}{2\pi i} \int_{|w|=\gamma} \frac{\exp(kw)}{w^{k+1}} \frac{1}{1-w} \; dw
Extracting the coefficient we get
k! \sum_{v=0}^k [w^v] \exp(kw) [w^{k-v}] \frac{1}{1-w} = k! \sum_{v=0}^k \frac{k^v}{v!}
as claimed.
Remark. This all looks very familiar but I am unable to locate the
duplicate among my papers at this time.