How to solve for $x$ in the following equation? Solve for $x \in \Bbb R$, given $$(x+m)^m-\binom{m}{1} (x+m-1)^m+\binom{m}{2}(x+m-2)^m + \cdots  + (-1)^m \binom{m}{m} x^m=m!$$ 
 A: This is an Identity! Therefore, this is true $ \forall x \in \mathbb R$.  I came to know about this today, that's why I am sharing this with you all. Let's prove this : 
Let's find coefficient of $y^m$ in $e^{xy}(e^y-1)^m$.
Expanding $(e^y-1)^m$ through binomial in LHS and Taylor series of $e^y$ in RHS, we've $$\text{Coefficient of } y^m \; \text{in} \; e^{xy}\left(e^{my}-\binom{m}{1} e^{(m-1)y}+\binom{m}{2} e^{(m-2)y}-\ldots (-1)^m\binom mm\right)=\text{Coefficient of } y^m \; \text{in} \; e^{xy} \left [ \left(1+y + \frac{y^2}{2!} +\ldots \right)-1\right]^m $$
Cancelling out $e^{xy}$ both the sides we get -
$$\boxed{\color{blue}{(x+m)^m-\binom{m}{1} (x+m-1)^m+\binom{m}{2}(x+m-2)^m \ldots (-1)^m \binom{m}{m} x^m=m!}}$$ 
A: Here is a slightly different variation  on the proof that was given by
OP. Introduce
$$P(x) = \sum_{q=0}^m (-1)^q {m\choose q} (x+m-q)^m.$$
We evaluate
$$[x^p] P(x)$$
where clearly $0\le p\le m.$ We have the claim if this is zero
for $p\gt 0$ and $m!$ for $p=0.$
Extracting the coefficient yields
$$\sum_{q=0}^m (-1)^q {m\choose q} {m\choose p} (m-q)^{m-p}
=  {m\choose p}
\sum_{q=0}^m (-1)^q {m\choose q} (m-q)^{m-p}
\\ =  {m\choose p}
\sum_{q=0}^m (-1)^{m-q} {m\choose q} q^{m-p}
\\ =  {m\choose p} (m-p)! [z^{m-p}]     
\sum_{q=0}^m (-1)^{m-q} {m\choose q} \exp(qz)
\\ =  {m\choose p} (m-p)! [z^{m-p}]  (\exp(z)-1)^m.$$
Note however that  $\exp(z)-1 = z + \cdots$ and  hence when $m-p\lt m$
we get a zero value  because the coefficient vanishes.  This condition
is equivalent to  $p\gt 0$ as required.  On the  other hand when $p=0$
we get
$${m\choose 0} m! [z^m] (\exp(z)-1)^m = m!$$
as claimed.
