The Wronskian of $n$ functions $y_1(x),\ldots,y_n(x)$ that are $\mathcal{C}^{n-1}$ in some interval $I$ is defined to be the determinant of the matrix
\begin{pmatrix} y_1(x) & ⋯ & y_n(x) \\ y_1'(x) & \ddots & y_n'(x) \\ \vdots & & \vdots \\ y_1^{(n-1)}(x) & \cdots & y_n^{(n-1)}(x) \end{pmatrix}
and is denoted $W(y_1,\ldots,y_n)(x)$.
I want to show that if $W(y_1,\ldots,y_n)(x)=0$ in $I$, but $W(y_1,\ldots,y_{n-1})(x)\ne 0$ for all $x ∈ I$, then the functions $y_1,\ldots,y_n$ are linearly dependent in $I$.
I want to say that if $y$ is an arbitrary function that is $\mathcal{C}^{n-1}$ and suppose that $W(y_1,\ldots,y_{n-1},y) = 0$, then by using cofactors on the last column I can see that the coefficient of $y^{(n-1)}(x)$ is non-zero since it's $W(y_1,\ldots,y_{n-1})(x)≠ 0$, so this is a linear homogeneous differential equation of order $(n-1)$ in $y$, call it $(\star)$ (it has a lot of determinants as its coefficients so I don't want to write it out).
Now I want to say that $y_1,\ldots,y_n$ solve $(\star)$. I can clearly see that $y_n$ at least solves it since $W(y_1,\ldots,y_n)(x)=0$, and in the computation of the determinant using cofactors with the last column I see $y_n$ satisfies $(\star)$ with the exact same coefficients. However, I can't see why $y_1,\ldots,y_{n-1}$ satisfy $(\star)$ or if this is even true.
(The thought is, if I can prove the above, then the claim follows from the fact that the space of all solutions of a linear homogeneous diff eq of order $(n-1)$ is indeed $(n-1)$. )
Going the other way, from the first $n-1$ rows of the matrix and the given determinants, you find that there exist factor functions $u_1(t),...,u_{n-1}(t)$ so that $$ y_n^{(k)}(t)=u_1(t)\,y_1^{(k)}(t)+...+u_{n-1}(t)\,y_{n-1}^{(k)}(t),\qquad k=0,1,...,n-2 $$ Now the question remains what you define as linear dependence of functions.