Translation of a polynomial I am given a complex polynomial $p(z) = a_0 + a_1z + \cdots + a_nz^n$, with $a_j \in \Bbb C$ for all $j$. Then, we fix $z_0 \in \Bbb C$ and define $P(z) = p(z+z_0)$, and I must prove that $P$ is also a polynomial. By doing some examples, comparing with the Pascal triangle, doing case $n = 4$, looking for patterns and such, I can conjecture that: $$P(z) = \sum_{k = 0}^nb_kz^k, \quad \text{where} \quad b_k = \sum_{j=0}^{n-k} {k+j \choose j}a_{k+j}z_0^{j}.$$
But I'm having trouble proving this. If I go for straight for the definition: $$\begin{align} P(z)=p(z+z_0) &= \sum_{k=0}^na_k(z+z_0)^k \\ &= \sum_{k=0}^n a_k\sum_{j=0}^k{k \choose j}z_0^{k-j}z^j  \\ &= \sum_{k=0}^n \left(\sum_{j=0}^k{k \choose j}a_kz_0^{k-j}z^j\right),\end{align}$$ and I'm stuck. I think that there must be some clever  relabeling of indices here, but I can't see it. Can someone help me please? Thanks.
(I should also add that the approach to this problem is preferred to be elementary. I have thought in the lines of Taylor expanding $p$ in some way, but we should not kill an ant with a cannon.)
 A: We can group terms by powers of $z$ as
\begin{align*}
P(z) &= \sum_{k=0}^n \left( \sum_{j=0}^{k} {k \choose j} a_k z_0^{n-j} z^j \right)\\
&= \left( {0 \choose 0} a_0 z_0^{n-0} + {1 \choose 0} a_1 z_0^{n-1} + \ldots + {n \choose 0} a_n z_0^{n-n} \right) z^0 \\
&\hspace{10mm} + \left( {1 \choose 1} a_1 z_0^{n-1} + \ldots + {n \choose 1} a_n z_0^{n-n} \right) z^1 + \ldots \\
&= \sum_{k=0}^n \left( \sum_{j=k}^n {j \choose k} a_j z_0^{n-(n-(j-k))} \right) z^k
\\
&= \sum_{k=0}^n \left( \sum_{i=0}^{n-k} {i + k \choose k} a_{i+k} z_0^{i} \right) z^k,
\end{align*}
where the last equality follows by making the substitution $i = j-k$. 
A: A slightly different approach:
Let $e_k(z) = z^k$, $f_k(z) = (z+z_0)^k$, and show that with
$E_n = \operatorname{sp} \{ e_k \}_{k=0}^n$, $ F_n = \operatorname{sp} \{ f_k \}_{k=0}^n$ we have $E_n = F_n$.
It is clear that $E_0 = F_0$. Suppose $E_n = F_n$. We have
$f_{n+1}(z) = z (z+z_0)^n + z_0 (z+z_0)^n = z f_n(z)+ z_0 f_n(z)$, and since
$f_n \in E_n$, we have $f_{n+1} \in E_{n+1}$. It follows that $F_{n+1} \subset E_{n+1}$.
Similarly, 
$e_{n+1}(z) = (z+z_0)z^n - z_0 z^n = (z+z_0)e_n(z) - z_0 e_n(z)$, and
since $e_n \in F_n$ we have $e_{n+1} \in F_{n+1}$. It follows that $E_{n+1} \subset F_{n+1}$.
Addendum: Actually, the above can be shortened using the binomial theorem.
Since $f_k = \sum_{i=0}^k \binom{k}{i} z_0^{k-i} e_i$, we have $f_k \in E_n$ for all $k \le n$ and so $F_n \subset E_n$.
Since $e_k = \sum_{i=0}^k \binom{k}{i} (-z_0)^{k-i} f_i$, we have $e_k \in F_n$ for all $k \le n$ and so $E_n \subset F_n$.
A: I'll add a way to do it by induction. First, I'll modify your notation slightly to make it more induction-hypothesis-friendly: $b_k^{(n)} = \displaystyle \sum_{j=0}^{n-k}{k+j \choose k}a_{k+j}z_0^j$
I'll limit myself to the inductive step, $p$ is now a polynomial of degree $n+1$. 
$$P(Z)=a_{n+1}(z+z_0)^{n+1}+\sum_{k=0}^{n}a_k(z+z_0)^k=a_{n+1}\sum_{k=0}^{n}{n+1 \choose k}z_0 ^{n+1-k}z^k+\sum_{k=0}^{n}b_k ^{(n)} z^k=\sum_{k=0}^{n}z^k\left ( a_{n+1}{n+1 \choose k}z_0 ^{n+1-k} + b_k ^{(n)} \right ) +a_{n+1}z^{n+1}$$
Now note that, by definition $a_{n+1}=b_{n+1} ^{(n+1)}$
Also: $a_{n+1}{n+1 \choose k}z_0 ^{n+1-k} + b_k ^{(n)}=a_{n+1}{n+1 \choose k}z_0 ^{n+1-k} + \displaystyle \sum_{j=0}^{n-k}{k+j \choose j}a_{k+j}z_0 ^j=a_{k+((n+1)-k)}{k+((n+1)-k) \choose k}z_0 ^{n+1-k}+\sum_{j=0}^{n-k}{k+j \choose j}a_{k+j}z_0 ^j=\sum_{j=0}^{n+1-k}{k+j \choose j}a_{k+j}z_0 ^j = b_{k}^{(n+1)}$
