Is $\sum_{p=0}^{k}(-1)^p\binom{k}{p}p^k=\sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+a)^k, a \in \mathbb{R}$ true? If so, how? I come here from this answer actually: https://math.stackexchange.com/a/3743805/765852
Due to re-indexing,
$$\sum_{p=1}^{k+1}(-1)^p\binom{k}{p-1}p^{k+1}$$
should equal
$$-\sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+1)^k,$$
yet it's written as
$$-\sum_{p=0}^{k}(-1)^p\binom{k}{p}p^k.$$ I tested some results with my calculator and it seems that this is not only true for $(p+1)^k$, but also for $(p+a)^k$, where $a \in \mathbb{R}$. Is this the case? If so, how would one go about proving this?
I actually tried my hand at proving this (it's just using simple algebra). However it requires a property which I also tried proving but I am not sure about its proof. I'm going to shamelessly plug that in here (it's open): How would I prove $\sum_{p=1}^{n}\binom{n}{p}p^m(-1)^p=0, m \in \mathbb{N}, 1\leq m < n $ via induction?
Ok here's the proof:
Statement:
$$\sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+a)^k=\sum_{p=0}^{k}(-1)^p\binom{k}{ p}p^k, a \in \mathbb{R}$$
Proof:
\begin{align*}
 \sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+a)^k &= \sum_{p=0}^{k}(-1)^p\binom{k}{p}(p^k+kp^{k-1}a+\binom{k}{2}p^{k-2}a^2+...kpa^{k-1}+a^k) \\
      &= \sum_{p=0}^{k}(-1)^p \binom{k}{p}p^k+ (-1)^p\binom{k}{p}kp^{k-1}a + (-1)^p\binom{k}{p} \binom{k}{2}p^{k-2}a^2... \\
      &= \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^k+ \sum_{p=0}^{k} (-1)^p\binom{k}{p}kp^{k-1}a+ \sum_{p=0}^{k} (-1)^p\binom{k}{p} \binom{k}{2}p^{k-2}a^2... \\
      &= \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^k+ka \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^{k-1}+a^2 \binom{k}{2} \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^{k-2}...       
        \end{align*}
Then, by the property linked above, it should be
$$\sum_{p=0}^{k} (-1)^p\binom{k}{p}p^k+0+0+....+0+ \sum_{p=0}^{k} (-1)^p\binom{k}{p}$$
Simply by the binomial theorem, the last term should also be zero. This completes the proof.
 A: Using
\begin{eqnarray*}
[x^k]: k! e^{(a+p)x} = (a+p)^k.
\end{eqnarray*}
We have
\begin{eqnarray*}
\sum_{p=0}^{k} (-1)^p \binom{k}{p}(a+p)^k &=& [x^k]: k! \sum_{p=0}^{k} (-1)^p \binom{k}{p} e^{ax} (e^x)^p \\ 
&=& [x^k]:k!  e^{ax} (1-e^x)^k= (-1)^k k! \\ 
\end{eqnarray*}
and so the sum is independent of $a$.
A: For any function $f$ defined on the real numbers, let $\def\D{\Delta}\D f$ denote the function defined by
$$
(\D f)(x)=f(x+1)-f(x)
$$
You can then iterate this construction. Let $\D^n f$ denote $\D(\D(\dots(\D f)\dots))$, with $n$ applications of $\D$. For example,
\begin{align}
(\D^2f)(x)
&=(\D f)(x+1)-(\D f)(x)
\\&=f(x+2)-2f(x+1)+f(x)\\
(\D^3 f)(x)&=f(x+3)-3f(x+2)+3f(x+1)-f(x)\\
(\D^4 f)(x)&=f(x+4)-4f(x+3)+6f(x+2)-4f(x+1)+f(x)
\end{align}
and so on. You might see Pascal's triangle emerging, and indeed, it is easy to prove by induction that
$$
(\D^k f)(x)=\sum_{p=0}^n (-1)^{k-p}\binom{k}pf(x+p)
$$
Now, letting $f(x)=x^k$, and factoring out $(-1)^k$, we get that
$$
(\D^k f)(x)=(-1)^p\sum_{p=0}^k (-1)^p\binom{k}p(p+x)^k\tag1
$$
You want to prove that the value of the last equation is independent of $x$. This is equivalent to saying that $\D^k f$ is a constant function. This is proven using the following observation:

For any polynomial $f$ of degree $n$, where $n>0$, $\D f$ is a polynomial of degree $n-1$.

Therefore, since $f(x)=x^k$ has degree $k$, it follows $\D^k f$ has degree zero, so is constant, so $(1)$ is independent of $x$.
