Formal Power Series as Linear Operators Let $t^k$ act as the $k$-th derivative operator on the set of polynomials.  So
$$t^k(x^n)=t^k x^n=(n)_kx^{n-k}$$
where $(n)_k=n(n-1)(n-2)...(n-k+1)$ is the falling factorial.  Then with a formal power series, $f(t)=\sum_{k\ge 0}a_k\frac{t^k}{k!}$, the linear operator $f(t)$ acts as such that
$$f(t)(x^n)=f(t)x^n=\sum_{k=0}^n\binom{n}{k}a_k x^{n-k}$$
Therefore, depending on the coefficients of the power series, we can get some interesting binomial identites.  For example, if $f(t)=e^{yt}$, since the coefficients $a_n=y^n$, we get
$$e^{yt}x^n=\sum_{k=0}^n\binom{n}{k}y^k x^{n-k}=(x+y)^n$$
by linearity, 
$$(e^{yt}-1)x^n=(x+y)^n-x^n=\sum_{k=1}^{n}\binom{n}{k}y^k x^{n-k}$$
and perhaps not as obvious
$$\left(\frac{e^{yt}-1}{t}\right)x^n=\int_{x}^{x+y}u^ndu$$
Now suppose that $f(t)=e^{yt}-1-yt$.  Then
$$(e^{yt}-1-yt)x^n=(x+y)^n-x^n-ynx^{n-1}=\sum_{k=2}^{n}\binom{n}{k}y^k x^{n-k}$$
Obviously there is a nice formed forward difference equation in the previous case that is not happening here.  But there is a relationship with subtracted terms of the binomial expansion.  What i would really like help understanding is whether or not a possible analogous integral representation exists for the following operator:
$$\left(\frac{e^{yt}-1-yt}{t^2}\right)x^n=\left(\sum_{k=0}^\infty\frac{y^{k+2}}{(k+2)(k+1)}\frac{t^k}{k!}\right)x^n=\sum_{k=0}^n\binom{n}{k}\frac{y^{k+2}}{(k+1)(k+2)}x^{n-k}$$  
$$=\sum_{k=0}^n\binom{n+2}{k+2}\frac{y^{k+2}}{(n+1)(n+2)}x^{n-k}=\frac{1}{(n+1)(n+2)}\sum_{k=2}^{n+2}\binom{n+2}{k}y^kx^{n+2-k}$$  
It is not as simple.  Clearly $\frac{d^2}{dx^2}\frac{x^{n+2}}{((n+2)(n+1)}$.  If I integrated below I think the math is correct
$$\int_x^{x+y}{\frac{u^{n+1}}{n+1}}du=\frac{1}{(n+1)(n+2)}\sum_{k=1}^{n+2}\binom{n+2}{k}y^kx^{n+2-k}$$  
Which is really close, but the lower bound on the summation is $1$, not $2$.  Does any one have any insight in how i can fix this, if possible?

Introductionary reference:

Operator Calculus

So why not replace $\,t\,$ by the differential operator $\,d/dx\,$ everywhere it occurs? Next reference:


Exponential of a function times derivative

Therefore we have in general and especially:
$$
e^{y\,d/dx} f(x) = f(x+y) \quad \Longrightarrow \quad e^{y\,d/dx} x^n = (x+y)^n
$$
Indeed the operator $1/(d/dx)$ is the inverse of differentiation, which is integration. So let's take a look at your perhaps not as obvious expression:
$$
\left(\frac{e^{y\,d/dx}-1}{d/dx}\right)x^n=\frac{1}{d/dx}\left(e^{y\,d/dx}-1\right)x^n\\
=\int\left[(x+y)^n - x^n\right] dx = \frac{(x+y)^{n+1}-x^{n+1}}{n+1} + C
$$
with $C$ an arbitrary constant. Evaluation is not ambiguous, because the operators $1/(d/dx)$ and $\left(e^{y\,d/dx}-1\right)$ do indeed commute. Which
means that we can do algebra with these operators as with common numbers. The same in, last but not least:
$$
\left(\frac{e^{y\,d/dx}-1-y\,d/dx}{(d/dx)^2}\right)x^n=\int\left\{\int\left[(x+y)^n-x^n\right]dx\right\}dx-\int y\,x^n\,dx
\\=\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} + Cx + D
$$
with $C$ and $D$ arbitrary constants. This result is deviant from your last formula. It is noticed that:
$$
\frac{d^2}{dx^2} \left[\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} + C\,x + D\right]
= (x+y)^n - x^n - y\frac{dx^n}{dx}
$$
EDIT. I almost forgot to mention the stuff where it's all about: the formal power series
(Wikipedia).
$$
(y+x)^{n}=\sum _{k=0}^{n}{n \choose k}y^{n-k}x^{k} \quad \Longrightarrow \\
\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} =\\
\frac{1}{(n+1)(n+2)}\left[\sum _{k=0}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k} - x^{n+2} - (n+2)y\,x^{n+1}\right]=\\
\frac{1}{(n+1)(n+2)}\left[\sum _{k=2}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k}
+ {n+2 \choose 0}y^0 x^{n+2} - x^{n+2} + {n+2 \choose 1}y^1 x^{n+1} - (n+2)y\,x^{n+1} \right]\\
=\frac{1}{(n+1)(n+2)}\left[\sum _{k=2}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k}\right]
$$
So here is your fix on the lower bound of the summation.
