$\newcommand\Ocal{\mathcal{O}}\newcommand\Hom{\mathop{\mathrm{Hom}}}\newcommand\Ext{\mathop{\mathrm{Ext}}\nolimits}$I think the following is a rather clumsy way of proving it. It is an induction on two variables, but probably there are better ways to set up the induction. The idea is to prove it for

- $i$ between 0 and $n$, $j=0$
- $i=0$, $j$ between 0 and $n$
- show that the cases $(i,j)$ and $(i-1,j-1)$ are isomorphic

I will denote the sequence that Beilinson suggests to do the induction with by $(\#_i)$.

**First part** The case $i=0$ is clear. To see that $\Hom(\Omega^i(i),\Ocal)=\bigwedge^iV^*$ for $i\geq 1$ we apply $\Hom(-,\Ocal)$ to $(\#_i)$. One gets a four-term exact sequence

$$0\to\Hom(\Omega^{i-1}(i),\Ocal)\to\Hom(\bigwedge\nolimits^iV\otimes\Ocal,\Ocal)\to\Hom(\Omega^i(i),\Ocal)\to\Ext^1(\Omega^{i-1}(i),\Ocal)\to 0$$

and isomorphisms $\Ext^l(\Omega^i(i),\Ocal)\cong\Ext^{l+1}(\Omega^{i-1}(i),\Ocal)$ for $l\geq 1$. The second term in the sequence is isomorphic to $\bigwedge^i V^*$, which is what we are after. So we would like to have that $\Ext^l(\Omega^{i-1}(i),\Ocal)=0$ for all $l$.

**Lemma** $\Ext^l(\Omega^{i-1}(i),\Ocal)=0$ for all $l$, and $i=1,\dotsc,n$.

*Proof* The case $i=1$ is clear. So assume that $i\geq 2$, and now we apply $\Hom(-,\Ocal(-1))$ to $(\#_{i-1})$. In the long exact sequence we get we see that $\Ext^l(\bigwedge\nolimits^{i-1}V^*\otimes\Ocal,\Ocal(-1))=0$ for all $l$. But then we apply the induction hypothesis, and conclude that everything is zero. $\square$.

So we get the desired isomorphism in this case.

**Second part** The case $j=0$ is clear. The case $j=n$ is also clear, because $\Omega^n(n)=\Ocal(-1)$. Then we rewrite the short exact sequence $(\#_{j+1})$ as

$$0\to\Omega^{j+1}(j)\to\bigwedge\nolimits^{j+1}\otimes\Ocal(-1)\to\Omega^j(j)\to 0$$

and then applying $\Hom(\Ocal,-)$ to this modified sequence gives a long exact sequence where $\Ext^l(\Ocal,\bigwedge\nolimits^{j+1}V\otimes\Ocal(-1))=0$, and similar to the previous lemma we can sandwich the desired term between zeroes.

**Third part** By now the ideas should be clear. Assume that $i\geq 1$ and $j\geq 1$. Apply $\Hom(-,\Omega^j(j))$ to $(\#_i)$. By the second part we see that the second (and fifth, etc.) term vanishes. We get another $\Ext^l=\Ext^{l+1}$ isomorphism.

Now apply $\Hom(\Omega^{i-1}(i),-)$ to $(\#_j)$. We get yet another long exact sequence, where by induction the third term $\Hom(\Omega^{i-1}(i),\Omega^{j-1}(j))$ is isomorphic to $\bigwedge\nolimits^{i-j}V^*$ (and higher $\Ext^l$ vanish).

Now we do yet another induction on as in the lemma, to see that the second (and fifth, etc.) term vanishes. Hence tracing back all the isomorphisms we see that

\begin{align}
\bigwedge\nolimits^{i-j}V^*
&\cong\Hom(\Omega^{i-1}(i-1),\Omega^{j-1}(j-1)) \\
&\cong\Hom(\Omega^{i-1}(i),\Omega^{j-1}(j)) \\
&\cong\Ext^1(\Omega^{i-1}(i),\Omega^j(j)) \\
&\cong\Hom(\Omega^i(i),\Omega^j(j))
\end{align}

as desired. Hurray!

The geometric intuition should correspond to its interpretation as a dual exceptional collection, but I don't see a way to make this into an actual explanation. The above is just a proof by tedious and boring computation. I hope I didn't make mistakes.