Show that $\alpha$ is zero with multiplicity $k$ iff derivatives at $\alpha$ are zero 
Show that $\alpha$ is a zero of multiplicity $k$ of the polynomial $P(z)$ if and only if $$P(\alpha) = P'(\alpha) = \ldots = P^{(k-1)}(\alpha) = 0$$ and $P^{(k)}(\alpha) \neq 0$.

So by definition of multiplicity, $P(z)=(z-\alpha)^k Q(z)$, where $Q(\alpha)\neq 0.$ Then we take derivatives up until the $k$-th derivative when the $(z-\alpha)$ term goes away. I took the first derivative directly and renamed it as a polynomial $R(z)$, where $R(\alpha)\neq 0$. It depends on $Q(z)$. The following derivative formulas get messy.
I wonder if there is a simple formula or clearer proof by induction. Thanks!
 A: We prove by induction on $n$ that $\alpha$ is a root of $P(z)=0$ of multiplicity $n$ if and only if
$$P(\alpha)=P'(\alpha)=\cdots =P^{(n-1)}(\alpha)=0\quad\text{and}\quad P^{(n)}(\alpha)\neq 0.$$
The case $n=0$ holds trivially, since the $0$-th derivative of $P$ is $P$.
Suppose that the result holds when $n=k$. We show that the result holds when $n=k+1$.  We need an auxiliary lemma. 
Suppose that $P(z)=(z-\alpha)Q(z)$. Then $P'(z)=Q(z)+(z-\alpha)Q'(z)$, and therefore $P''(z)=2Q'(z)+(z-\alpha)Q''(z)$, and therefore $P^{(3)}(z)=3Q''(z)+(z-\alpha)Q^{(3)}(z)$, and therefore
$P^{(4)}(z)=4Q^{(3)}(z)+(z-\alpha)Q^{(4)}(z)$, and so on. In general,
Lemma.
$\:P^{(j)}(z)=jQ^{(j-1)}(z)+(z-\alpha)Q^{(j)}(z).$
In principle, we should prove the above lemma by induction. However, the pattern is so obvious that writing out the details is a waste of time. By setting $z=\alpha$ in the above lemma, we obtain
 $$P^{(j)}(\alpha)=j\,Q^{(j-1)}(\alpha).\qquad(\ast)$$
Now suppose that $\alpha$ is a root of $P(z)$ of multiplicity $k+1$. Then the polynomial $z-\alpha$ divides $P(z)$, by the Factor Theorem. Let $P(z)=(z-\alpha)Q(z)$. Then $\alpha$ is a root of $Q(z)$ of multiplicity $k$.  It follows by the induction hypothesis that $Q^{(j)}(\alpha)=0$ for all $j \le k-1$.  It then follows from $(\ast)$ that $P^{(j)}(\alpha)=0$ for all $j \le k$.   We must also show that $P^{(k+1)}(\alpha)\ne 0$. This again follows from $(\ast)$, since by the induction hypothesis, $Q^{(k)}(\alpha)\ne 0$.
Next  we prove the converse assertion: If $P^{(j)}(\alpha)=0$ for all $j \le k$ and $P^{(k+1)}(\alpha)\ne 0$, then $\alpha$ is a root of $P(z)$ of multiplicity $k+1$. By putting $j=0$, we know that $\alpha$ is a root of $P(z)$.  So by the Factor Theorem, $z-\alpha$ divides $P(z)$. Let $P(z)=(z-\alpha)Q(z)$.   
By $(\ast)$,  since $P^{(j)}(\alpha)=0$ for all $j\le k$, we conclude that $jQ^{(j-1)}=0$ for all $j \le k$, and therefore $Q^{(i)}(\alpha)=0$ for all $i \le k-1$. Moreover, since 
$P^{(k+1)}(\alpha)\ne 0$, we conclude from $(\ast)$ that $Q^{(k)}(\alpha)\ne 0$. It follows by the induction hypothesis that $\alpha$ is a root of $Q(z)$ of multiplicity $k$, meaning that $(z-\alpha)^k$ divides $Q(z)$ but no higher power of $z-\alpha$ does. Hence $(z-\alpha)^{k+1}$ divides $P(z)$ but no higher power of $z-\alpha$ does. Thus  $\alpha$ is a root of $P(z)$ of multiplicity $k+1$.  
Remark: We wrote out the proof in great detail. There is much less to it than meets the eye!  Apart from the Factor Theorem,  the key fact, indeed essentially the only fact, that we needed is $(\ast)$.
