By induction on $k$ one can easily reduce the question to $k=2$.

**Claim** If $\mathscr I, \mathscr J\subseteq \mathscr O_X$ are two ideal sheaves such that $\mathscr I +\mathscr J =\mathscr O_X$, then $\mathscr I\mathscr J=\mathscr I\cap \mathscr J$.

**Proof** 
First, observe that the condition implies that 

`\begin{equation}\tag{1}
\mathscr O_X/(\mathscr{I\cap J}) \simeq (\mathscr O_X/\mathscr I) \oplus (\mathscr O_X/\mathscr J). 
\end{equation}`

Second, observe that since $\mathrm{supp} \mathscr I\cap \mathrm{supp}\mathscr J=\emptyset$, we have that
`\begin{equation}\tag{2}
\mathscr{I/(IJ)}\simeq \mathscr I\otimes (\mathscr O_X/\mathscr J) \simeq \mathscr O_X\otimes (\mathscr O_X/\mathscr J) \simeq \mathscr O_X/\mathscr J
\end{equation}`

Next consider the following commutative diagram with the "obvious" natural maps

`\begin{align}
0 &\to & \mathscr I/(\mathscr I\mathscr J) & \to  &\mathscr O_X/(\mathscr{IJ}) & \to & \mathscr O_X/\mathscr I  & \to & 0 \\
 & & \downarrow\qquad\  & & \downarrow\qquad\  & & \downarrow\quad\  & \\
0 &\to & \mathscr I/(\mathscr I\cap\mathscr J) & \to &\mathscr O_X/(\mathscr{I\cap J}) & \to & \mathscr O_X/\mathscr I & \to & 0 \\
&&& & \downarrow\qquad\  & \\
&&& & \mathscr O_X/\mathscr J.
\end{align}`

By $(1)$ the induced map $\mathscr{I/(I\cap J)}\to \mathscr O_X/\mathscr J$ is an isomorphism of $\mathscr O_X$-modules and by $(2)$ the induced map $\mathscr{I/(IJ)}\to \mathscr O_X/\mathscr J$ is also an isomorphism of $\mathscr O_X$-modules.

This implies that the natural map
$$
\mathscr{I/(IJ)}\to \mathscr I/\mathscr{(I\cap J)}
$$
is an isomorphism and hence $\mathscr{IJ=I\cap J}$.