If $\sum_{i=1}^r\lambda_i^n=0$ for every $n\in\mathbb N$, then $\lambda_i=0\;\forall i=1,2,...,r$ 
If $\sum_{i=1}^r\lambda_i^n=0$ for every $n\in\mathbb N$, then show that $\lambda_i=0\space\forall i=1,2,...,r$.

I have done it assuming $\lambda_1,\lambda_2,...,\lambda_r$ are roots of an $r$-degree polynomial and then showing that the given condition implies that all coefficients of this polynomial, except the one of $x^r$, is $0$, giving us the polynomial $f(x)=x^r$ whose roots are all $0$, hence $\lambda_i=0$ $\forall i=1,2,...,r$.
However, I got the following solution from a friend of mine:

Consider the vectors $a_i=(\lambda_1^i,...,\lambda_r^i)$ for each $i\in\mathbb N$. Then, there are elements of $\mathbb C^r$, which can have $k$ basis elements for some finite $k$. Now, the dot product of any two of these $a_i$ yields $0$, implying that these vectors are linearly independent, so infinitely many linearly independent vectors have been achieved. This is a contradiction.

Is this correct?  I think there's a problem with this solution. I don't think dot product is a valid inner product in $\mathbb C^r$, because it does not give a norm, while every inner product should give a norm. Here $<x,x>=0$ does not necessarily mean $x=0$ where $x\in\mathbb C^r$.
So the orthogonality argument w.r.t. "dot product" is rendered meaningless here.
 A: Your friend's argument fails as stated in some comments and answers.  There is still a linear-algebraic argument.
Suppose that $\lambda_1$, $\lambda_2$, $\ldots$, $\lambda_k$ are all the pairwise distinct nonzero elements of the collection $\left(\lambda_i\right)_{i=1}^r$ (where we have reordered the entries, if necessary).  Let $m_0$ denote the number of zero entries of $\left(\lambda_i\right)_{i=1}^r$.   Assume that $\lambda_i$ occurs with multiplicity $m_i$ for $i=1,2,\ldots,k$ (so that $\sum_{i=0}^k\,m_i=r$).  We claim that $k=0$.  That is, $\lambda_i=0$ for all $i=1,2,\ldots,r$. 
Suppose contrary that $k>0$.  Let $\mathrm{V}$ be the modified Vandemonde matrix $\left[\lambda_i^j\right]_{i,j\in\{1,2,\ldots,k\}}$.  Then, $\mathrm{Vx}=\boldsymbol{0}$, where $\mathrm{x}:=\left(m_1,m_2,\ldots,m_k\right)$.  Note that $\mathbf{V}$ is invertible because $$\det(\mathbf{V})=\left(\prod_{i=1}^k\,\lambda_i\right)\left(\prod_{1\leq i<j\leq k}\,\left(\lambda_i-\lambda_j\right)\right)\neq 0\,.$$
Thus, $\mathbf{x}=\boldsymbol{0}$, but each $m_i$ for $i=1,2,\ldots,k$ must be a positive integer.  We have a contradiction, and so $k=0$.  The proof is now complete.
In general we have the following statement.

Let $r$ be a positive integer.  Suppose that $\lambda_1,\lambda_2,\ldots,\lambda_r$ are pairwise distinct elements of a field $\mathbb{K}$ such that there are $\alpha_1,\alpha_2,\ldots,\alpha_r\in \mathbb{K}\setminus\{0\}$ for which $\sum_{i=1}^r\,\alpha_i\lambda_i^n=0$ holds for each positive integer $n$.  Then, $r=1$ and $\lambda_1=0$.

A: There is an argument similar to the one you suggest you have used. Assume that none of the $\lambda_i$ is equal to $0$ (by throwing away the zeros, which obviously make no difference to the condition). Now, consider an operator $M$ whose eigenvalues are the $\lambda_i.$ By Cayley-Hamilton, this satisfies its minimal polynomial, so $p(M) = 0.$ By taking traces of both sides, you see that the constant term is zero (since the trace of $I$ is not zero). But that means that $0$ is an eigenvalue of $M,$ contrary to the assumption.
A: You are misunderstanding what the inner product on $\mathbb{C}^r$ is. You should have
$$\langle x,x\rangle=\overline{x}^T\cdot x$$
which is nondegenerate.
