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.

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 $a$.
