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Here is a solution in the $C^1$ case [but see upd]. Suppose the vectors $F'(t_1),\ldots, F'(t_n)$ are linearly independent for all $0\leq t_1< \cdots < t_n\leq 1$. Let $L(t_1,\ldots,t_{n-1})$ be vector space spanned by the first $n-1$ of these. Let $$t^i=(t_1^i,\ldots,t_{n-1}^i),0\leq t^i_1< \cdots < t^i_{n-1}$$ t^i_{n-1}\leq 1$$be a sequence such that t^i_{n-1}\to 0 as i\to \infty. Since the space of all vector hyperplanes in \mathbf{R}^n is compact, we can assume the sequence L(t^i) has a limit L. Claim: for any s,t such that 0< s < t\leq 1 the hyperplane L does not separate F'(s) and F'(t). Proof of the claim: if L does, then for all sufficiently large i the hyperplane L(t^i) also separates F'(s) and F'(t). Choose i so that moreover t^i_{n-1} < s. Then the determinants of the matrices (F'(t^i_1),\ldots, F'(t^i_{n-1}),F'(s)) and (F'(t^i_1),\ldots, F'(t^i_{n-1}),F'(t)) have different signs. This is impossible, since F' is continuous and the determinant is never zero. Claim is proven. Now one of the two things can happen: either all F'(t),0 \leq t\leq 1 are in L, in which case F'(t_1),\ldots, F'(t_n) are linearly dependent for all t_1,\ldots, t_n, or there is one a F'(t) t such that for l(F'(t))\neq 0 where l is a linear equation defining L. Say, l(F'(t)) > 0. Then l(F(0)) > < l(F(1)), so we can't have F(0)=F(1). upd: here is how one can take care of the case when F' is not assumed continuous. Basically, the only thing that changes is the proof of the claim; the claim itself remains the same except that we assume t < 1. Choose i as above and set g(x),s\leq x\leq t to be the determinant of (F'(t^i_1),\ldots,F'(t^i_{n-1}),F(x)). This function is differentiable and we have g'(x)=det(F'(t^i_1),\ldots,F'(t^i_{n-1}),F'(x)). So g'(s) and g'(t) have different signs. The claim follows now from the following statement, which is a consequence of the classical Rolle's theorem: if f:[a,b]\to\mathbf{R} is differentiable at each point of [a,b] and f'(a) and f'(b) have different signs, then there is an x\in (a,b) such that f'(x)=0. Then we deduce from the claim that all F'(t),0 < t < 1 are in the same half-space with respect to L. This suffices. 2 added the case when the derivative is not assumed continuous Here is a solution in the C^1 case. Suppose the vectors F'(t_1),\ldots, F'(t_n) are linearly independent for all 0\leq t_1< \cdots < t_n\leq 1. Let L(t_1,\ldots,t_{n-1}) be vector space spanned by the first n-1 of these. Let$$t^i=(t_1^i,\ldots,t_{n-1}^i),0\leq t^i_1< \cdots < t^i_{n-1}$$be a sequence such that t^i_{n-1}\to 0 as i\to \infty. Since the space of all vector hyperplanes in \mathbf{R}^n is compact, we can assume the sequence L(t^i) has a limit L. Claim: for any s,t such that 0< s < t\leq 1 the hyperplane L does not separate F'(s) and F'(t). Proof of the claim: if L does, then for all sufficiently large i the hyperplane L(t^i) also separates F'(s) and F'(t). Choose i so that moreover t^i_{n-1} < s. Then the determinants of the matrices (F'(t^i_1),\ldots, F'(t^i_{n-1}),F'(s)) and (F'(t^i_1),\ldots, F'(t^i_{n-1}),F'(t)) have different signs. This is impossible, since F' is continuous and the determinant is never zero. Claim is proven. Now one of the two things can happen: either all F'(t),0 \leq t\leq 1 are in L, in which case F'(t_1),\ldots, F'(t_n) are linearly dependent for all t_1,\ldots, t_n, or there is one F'(t) such that for l(F'(t))\neq 0 where l is a linear eqation equation defining L. Say, l(F'(t)) > 0. Then l(F(0)) > l(F(1)), so we can't have F(0)=F(1). upd: here is how one can take care of the case when F' is not assumed continuous. Basically, the only thing that changes is the proof of the claim; the claim itself remains the same except that we assume t < 1. Choose i as above and set g(x),s\leq x\leq t to be the determinant of (F'(t^i_1),\ldots,F'(t^i_{n-1}),F(x)). This function is differentiable and we have g'(x)=det(F'(t^i_1),\ldots,F'(t^i_{n-1}),F'(x)). So g'(s) and g'(t) have different signs. The claim follows now from the following statement, which is a consequence of the classical Rolle's theorem: if f:[a,b]\to\mathbf{R} is differentiable at each point of [a,b] and f'(a) and f'(b) have different signs, then there is an x\in (a,b) such that f'(x)=0. Then we deduce from the claim that all F'(t),0 < t < 1 are in the same half-space with respect to L. This suffices. 1 Here is a solution in the C^1 case. Suppose the vectors F'(t_1),\ldots, F'(t_n) are linearly independent for all 0\leq t_1< \cdots < t_n\leq 1. Let L(t_1,\ldots,t_{n-1}) be vector space spanned by the first n-1 of these. Let$$t^i=(t_1^i,\ldots,t_{n-1}^i),0\leq t^i_1< \cdots < t^i_{n-1} be a sequence such that $t^i_{n-1}\to 0$ as $i\to \infty$. Since the space of all vector hyperplanes in $\mathbf{R}^n$ is compact, we can assume the sequence $L(t^i)$ has a limit $L$.
Claim: for any $s,t$ such that $0< s < t\leq 1$ the hyperplane $L$ does not separate $F'(s)$ and $F'(t)$.
Proof of the claim: if $L$ does, then for all sufficiently large $i$ the hyperplane $L(t^i)$ also separates $F'(s)$ and $F'(t)$. Choose $i$ so that moreover $t^i_{n-1} < s$. Then the determinants of the matrices $(F'(t^i_1),\ldots, F'(t^i_{n-1}),F'(s))$ and $(F'(t^i_1),\ldots, F'(t^i_{n-1}),F'(t))$ have different signs. This is impossible, since $F'$ is continuous and the determinant is never zero. Claim is proven.
Now one of the two things can happen: either all $F'(t),0 \leq t\leq 1$ are in $L$, in which case $F'(t_1),\ldots, F'(t_n)$ are linearly dependent for all $t_1,\ldots, t_n$, or there is one $F'(t)$ such that for $l(F'(t))\neq 0$ where $l$ is a linear eqation defining $L$. Say, $l(F'(t)) > 0$. Then $l(F(0)) > l(F(1))$, so we can't have $F(0)=F(1)$.