On the proof of a result of Bayer and Stillman I'm reading through the paper A criterion for detecting m-regularity of Bayer and Stillmann and came across a proof, where I don't understand an implication.
The following things may need to be mentioned:

*

*$S = k[x_1,\ldots,x_n]$, $I \subset S$ is a homogeneous ideal and $M$ a graded $S$-module

*$\mathfrak{m}$ = $(x_1,\ldots,x_n)$ denotes the irrelevant maximal ideal of $S$

*$H_{\mathfrak{m}}^{i}(M)_d$ will denote the degree $d$ part of the $i$-th local cohomology group of $M$

*$I$ is $m$-regular if $H_{\mathfrak{m}}^{i}(I)_d = 0 $ for all $i$ and $d \geq m-i+1$.

*$I$ is $m$-saturated if $I^{\textit{sat}}_d = I_d$ for all $d \geq m$.

*$h \in S$ is called generic for $I$ if $h$ is not a zero-divisor on $S/I^{\textit{sat}}$
It concerns the following Lemma:

Lemma 1.8: Let $I \subset S$ be an ideal, and suppose $h\in S_1$ is generic for $I$. TFAE:
(a) $I$ is $m$-regular
(b) $I$ is $m$-saturated, and $(I,h)$ is $m$-regular

The proof of the direction $(a)\implies(b)$ looks like this:

Proof.
Since $I$ is $m$-regular, it follows that $I$ is $m$-saturated by a previous remark (1.3 in paper). (Thus, it remains to be shown that $(I,h)$ is $m$-regular). 
Let $Q = (I:h)/I$ to get an exact sequence
\begin{equation}
0 \rightarrow I \rightarrow (I:h) \rightarrow Q \rightarrow 0.
\end{equation}
Since $I$ is $m$-saturated and $h$ generic for $I$, by a previous Lemma (1.6 in the paper) it follows that $I_d = (I:h)_d$ for all $d \geq m$ and therefore $\dim(Q) = 0.$ Thus, $H_{\mathfrak{m}}^{i}(Q) = 0$ for $i \neq 0$, and $H_{\mathfrak{m}}^{0}(Q) = Q$. 
Thus, by the long exact sequence for local cohomology we obtain
\begin{equation}
\tag{$\star$} H_{\mathfrak{m}}^{i}(I)_d \cong H_{\mathfrak{m}}^{i}((I:h))_d \text{ for } d \geq m -i+1 \text{ and all } i.
\end{equation}
Considering the exact sequence
\begin{equation}
0 \rightarrow I \cap (h) \rightarrow I \oplus (h) \rightarrow (I,h) \rightarrow 0.
\end{equation}
and the fact that $I\cap(h) = (I:h)h$, we get
\begin{equation}
0 \rightarrow (I:h)(-1) \rightarrow I\oplus(h) \rightarrow (I,h) \rightarrow 0,
\end{equation}
which leads to
\begin{equation}
 \tag{$\star \star$}H_{\mathfrak{m}}^{i}(I\oplus(h))_d \rightarrow H_{\mathfrak{m}}^{i}((I,h))_d\rightarrow H_{\mathfrak{m}}^{i+1}((I:h))_{d-1}.
\end{equation}

So far I think I can follow, but in the paper the proof gets concluded in the following way:

From $(\star\star)$ and the isomorphism $(\star)$ it follows that $(I,h)$ is $m$-regular.

Question:
If I have that $H_{\mathfrak{m}}^{i}((I,h))_d = 0 $ for all $i$ and $d \geq m-i+1$, then the conclusion follows by definition. However, I don't see how $(\star\star)$ and $(\star)$ imply this. 
Since $I$ is $m$-regular $H_{\mathfrak{m}}^{i}(I)_d = H_{\mathfrak{m}}^{i}((I:h))_d = 0 $ for all $i$ and $d \geq m-i+1$. But I think that is not enough to conclude the result.
What am I missing resp. why is the above enough to conclude the proof?
I'm rather unfamiliar with the concept of local cohomology, so the implication may be obvious, but I can't see it.
 A: Expanding on Youngsu's comment:
We have the isomorphism $$H_{\mathfrak{m}}^{i}(I)_d \cong H_{\mathfrak{m}}^{i}((I:h))_d \text{ for } d \geq m -i+1 \text{ and all } i\tag{$\star$}$$
and the exact sequence
$$H_{\mathfrak{m}}^{i}(I\oplus(h))_d \rightarrow H_{\mathfrak{m}}^{i}((I,h))_d\rightarrow H_{\mathfrak{m}}^{i+1}((I:h))_{d-1}.$$
We need to show that
$H_{\mathfrak{m}}^{i}((I,h))_d = 0 $ for $d \geq m-i+1$ and all $i$.
Notice that $H_{\mathfrak{m}}^{i}(I\oplus(h))\cong H_{\mathfrak{m}}^{i}(I)\oplus H_{\mathfrak{m}}^{i}((h))$ (see for instance [1]). Next, we have $(h)\cong S(-1)$ as $S$-modules.
So,
\begin{align}
H_{\mathfrak{m}}^{i}(I\oplus(h))_d
&=H_{\mathfrak{m}}^{i}(I)_d\oplus H_{\mathfrak{m}}^{i}(S(-1))_d\\
&=H_{\mathfrak{m}}^{i}(I)_d\oplus H_{\mathfrak{m}}^{i}(S)_{d-1}.\\
\end{align}
Recall that the local cohomologies of the polynomial ring $S=k[x_1,\ldots,x_n]$ is given by
$$H_{\mathfrak{m}}^{i}(S)_d=\begin{cases}\operatorname{Hom}_k(S_{-n-d},k),&i=n\\0,&\text{otherwise}\end{cases}$$
(see [Eisenbud, Cor.10.9]).
Thus $H_{\mathfrak{m}}^{i}(I\oplus(h))_d=0$ for  all $d\geq m-i+1$ and $i\neq n$. Suppose that $i=n$ and $d\geq m-n+1$. Then we have $$H_{\mathfrak{m}}^{n}(I\oplus(h))_d=H_{\mathfrak{m}}^{n}(I)_d\oplus H_{\mathfrak{m}}^{n}(S)_{d-1}=0\oplus \operatorname{Hom}_k(S_{-n-d+1},k).$$
Since $d\geq m-n+1$, we have $0\geq -m\geq -n-d+1$, so $S_{-n-d+1}=0$. Therefore, $H_{\mathfrak{m}}^{n}(I\oplus(h))_d=0$ for all $d\geq m-n+1$. Thus the left-hand term of the exact sequence vanishes for all $d\geq m-n+1$.
Finally, consider the right-hand term. For $d\geq m-i+1$, we have $d-1\geq m-(i+1)+1$, so  $H_{\mathfrak{m}}^{i+1}((I:h))_{d-1}$ is isomorphic to $H_{\mathfrak{m}}^{i+1}(I)_{d-1}$ by $(\star)$. As $I$ is $m$-regular, it follows that $H_{\mathfrak{m}}^{i+1}((I:h))_{d-1}=0$ for all $d\geq m-i+1$. Hence the middle term $H_{\mathfrak{m}}^{i}((I,h))_d$ also vanishes for $d \geq m-i+1$ and all $i$.
