Why does the Wronskian satisfy $W(yy_1,\ldots,yy_n)=y^n W(y_1,\ldots,y_n)$? The Wronskian of smooth functions $y_1,\ldots,y_n$ is defined by the determinant $$W(y_1,\ldots,y_n):=\det \left(y_i ^{(j)}\right).$$
It can be verified by a messy induction that the Wronskian satisfies the identity $$W(yy_1,\ldots,yy_n)=y^n W(y_1,\ldots,y_n)$$ for every smooth function $y$.

What is a conceptual proof of this fact?

Why should we "expect" the result to be true?
Note that it is not just multiplying each column by the same scalar. There are many cross-terms coming from the Leibnitz rule for derivatives.
For example, $n=2$ is the assertion $$yy_1(yy_2)'-yy_2(yy_1)'=y^2(y_1y_2'-y_2y_1').$$
The Wronskian has intuitive meaning as the volume spanned by solutions to an ODE of order $n$ after converting it to a system of first-order equations. 
The identity is true for all functions that are differentiable $n-1$ times, but giving a proof just in the case that they solve an ODE will also be great.
 A: This is a particular case of Theorem 1.50 in my Collected trivialities on algebra derivations (as of 18 October 2020), which says the following (up to different notations):

Theorem. Let $A$ be a commutative ring. Let $\delta : A \to A$ be a derivation. Let $a_1, a_2, \ldots, a_n \in A$ and $a \in A$. Then,
\begin{align}
W_\delta\left(aa_1, aa_2, \ldots, aa_n\right) = a^n W_\delta\left(a_1, a_2, \ldots, a_n\right) ,
\end{align}
where the $\delta$-Wronskian of an $n$-tuple $\left(b_1, b_2, \ldots, b_n\right) \in A^n$ is defined as the determinant of the $n\times n$-matrix
\begin{align}
\begin{pmatrix}
\delta^0\left(b_1\right) & \delta^1\left(b_1\right) & \cdots & \delta^{n-1}\left(b_1\right) \\
\delta^0\left(b_2\right) & \delta^1\left(b_2\right) & \cdots & \delta^{n-1}\left(b_2\right) \\
\vdots & \vdots & \ddots & \vdots \\
\delta^0\left(b_n\right) & \delta^1\left(b_n\right) & \cdots & \delta^{n-1}\left(b_n\right)
\end{pmatrix} .
\end{align}

To recover your claim from this theorem, let $A$ be the ring of smooth functions, and set $\delta = \dfrac{d}{dx}$ and $a_i = y_i$ and $a = y$.
The proof of the theorem follows the same plan as my comment: The trick is to write the matrix
\begin{align}
\begin{pmatrix}
\delta^0\left(aa_1\right) & \delta^1\left(aa_1\right) & \cdots & \delta^{n-1}\left(aa_1\right) \\
\delta^0\left(aa_2\right) & \delta^1\left(aa_2\right) & \cdots & \delta^{n-1}\left(aa_2\right) \\
\vdots & \vdots & \ddots & \vdots \\
\delta^0\left(aa_n\right) & \delta^1\left(aa_n\right) & \cdots & \delta^{n-1}\left(aa_n\right)
\end{pmatrix}
\end{align}
(whose determinant is $W_\delta\left(aa_1, aa_2, \ldots, aa_n\right)$) as a matrix product $BC$, where $B$ is the matrix
\begin{align}
\begin{pmatrix}
\delta^0\left(a_1\right) & \delta^1\left(a_1\right) & \cdots & \delta^{n-1}\left(a_1\right) \\
\delta^0\left(a_2\right) & \delta^1\left(a_2\right) & \cdots & \delta^{n-1}\left(a_2\right) \\
\vdots & \vdots & \ddots & \vdots \\
\delta^0\left(a_n\right) & \delta^1\left(a_n\right) & \cdots & \delta^{n-1}\left(a_n\right)
\end{pmatrix}
\end{align}
(whose determinant is $W_\delta\left(a_1, a_2, \ldots, a_n\right)$) and where $C$ is a certain upper-triangular matrix whose diagonal entries are $a, a, \ldots, a$. More precisely, $C$ is the upper-triangular matrix whose $\left(i,j\right)$-th entry is $\dbinom{j-1}{i-1} \delta^{i-j}\left(a\right)$ whenever $j \geq i$.
The same argument applies in the more general case when the functions $y_i$ are merely $n-1$-times differentiable (as opposed to smooth), even though the derivation $\delta = \dfrac{d}{dx}$ no longer literally exists as a derivation on a ring.
