A curve $(t,t^2,t^3,\ldots,t^n) \in\mathbb R^n$ is not contained in any hyperplane Let $C$ the curve given by $\gamma:\mathbb R\to\mathbb R^n$, $\gamma(t)=(t,t^2,t^3,\ldots,t^n)$. 


*

*Show that $C$ is contained in no hyperplane.


2.Find the tangent line and osculating plane at $(0,0,\ldots,0)$.
For (1) my idea is to find the curvature of $\gamma$. We have,
$$\gamma'(t)=(1,2t,3t^2,...,nt^{n-1})\quad \gamma''(t)=(0,2,6t,...,n(n-1)t^{n-2}).$$
Then $k(t)=||\gamma''(t)||=\sqrt[]{4+36t^2+\cdots+(n(n-1)t^{n-2})^2}>0$ so it is not contained in any hyperplane. Is it ok? If so, is there another way to solve this without using curvature?
To get the tangent line and osculating plane, I just take the binormal vector $B(t)=\gamma'(t)\times \gamma''(t)$ and them i get the plane.
Is it ok?
Thanks for your help. 
 A: Your solution involving the curvature doesn't work as positive curvature doesn't guarantee that the curve doesn't lie on a hyperplane if $n > 2$. For example, the curve $\gamma(t) = (\cos(t), \sin(t), 0)$ lies on the $xy$-plane but has curvature $k(t) = ||(-\cos(t), -\sin(t), 0)|| \equiv 1$. Also, your calculation of the curvature is not correct as the curve is not parametrized by arclength.
The correct generalization of the statement and a sufficiently nice curve lies on a line if and only if the curvature vanishes is that the curve lies on a hyperplane if and only if the last generalized curvature vanishes. For $n = 3$ this is the torsion of the curve, not the curvature.
Instead, you can argue directly. A hyperplane in $\mathbb{R}^n$ is described by a linear equation of the form
$$ a_1 x_1 + \dots + a_n x_n = c $$
for $ (0, \dots ,0) \neq (a_1, \dots, a_n) \in \mathbb{R}^n$ and $c \in \mathbb{R}$. If $\gamma(t)$ lies on such hyperplane, we must have
$$ a_1 t + \dots + a_n t^n = c $$
for all $t \in \mathbb{R}$. However, this is a polynomial equation in $t$ and so has at most $n$ solutions. Thus, we have shown that at most $n$ points on the image of $\gamma$ are contained in any given hyperplane.
For $(2)$, the tangent line at $t = 0$ will be $\{ \gamma(0) + t \dot{\gamma}(0) \, | \, t \in \mathbb{R} \}$. To find the osculating plane, again, you can't compute $\dot{\gamma}(0) \times \ddot{\gamma}(0)$ since the cross product of two vectors in $\mathbb{R}^n$ isn't even defined! Instead, the osculating plane is the plane spanned by the first two vectors in the generalized Frenet-Serret frame. Explicitly, 
$$e_1(t) = \frac{\dot{\gamma}(t)}{||\dot{\gamma}(t)||}, e_2(t) = \frac{\ddot{\gamma}(t) - \left< \ddot{\gamma}(t), e_1(t) \right>}{||\ddot{\gamma}(t) - \left< \ddot{\gamma}(t), e_1(t) \right>||} $$
and the osculating plane at $\gamma(t)$ is $\gamma(t) + \operatorname{span} \{ e_1(t), e_2(t) \}$.
A: In my opinion your idea does not work for a dimension greater than $2$.  
Hint for (1). Try to show that the polynomials $t,t^2,\dots,t^n$ are linearly independent functions (see https://en.wikipedia.org/wiki/Wronskian). 
Moreover, if there are constants $a_1,\dots, a_n$ such that
$$a_1 t+a_2 t+\dots +a_nt^n=0$$
for ALL $t\in \mathbb{R}$, what may we conclude?
A: Forget the curvature.
Let's take $n+1$ different points along the curve, the first of them being at $t=0$, just to make things look nice. Draw $n$ vectors from that point to all other points. Now, the non-(hyper)planarity of the curve is equivalent to rank of that system being $n$. In other words, these vectors must be a basis of all $\mathbb R^n$. The easiest way to verify that is to check the determinant of a matrix composed of these vectors - it must not be 0. But wait, that's the Vandermonde matrix, and as long as all points are different (which I said they are), $\det\ne0$.
A: If it is in some hyperplane then $\gamma(t)$ is between some two
hyperplanes
That is $$ |v=(v_1,\cdots ,v_n)|=1,\ (\gamma (t)-p)\cdot v \geq 0 $$
and
$$ (\gamma(t)-(-p))\cdot (-v) \geq 0 $$
Hence $$ C_1\leq \gamma(t)\cdot v \leq C_2 \ \ast$$
so that $$ \frac{C_1}{|t|^n }\leq \frac{\gamma(t)\cdot v}{|t|^n}
\leq \frac{C_2}{|t|^n} $$
When $|t|$ goes to $\infty$, then $v_n=0$ And in $\ast$ we divide by
$|t|^{n-1}$ so that we conclude $v_{n-1}=0$ Hence $v=0$ It is a
contradiction
