Linearly Dependent Sets and Linear Combinations (Proof Verification)

Question

I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.

The Prompt (from here):

for n > 1, (v₁, ..., v_n) is a linear dependent set if and only if one of the vector in the set v_i is a linear combination of the other n − 1 vectors.

My Proof:

Suppose (v₁,...,v_n) is a linearly dependent set. Then $\sum_{i=1}^n$ t_iv_i = 0 for some nonzero t ∈ Rⁿ. Let v_k be a vector s.t. t_k != 0. Then:

$\sum_{i=1}^n$ t_iv_i = 0

$\sum_{i=1}^{k-1}$ t_iv_i + t_kv_k + $\sum_{i=k+1}^n$ t_iv_i = 0

t_kv_k = -$\sum_{i=1}^{k-1}$ t_iv_i - $\sum_{i=k+1}^n$ t_iv_i

v_k = -($\sum_{i=1}^{k-1}$ t_iv_i + $\sum_{i=k+1}^n$ t_iv_i) / t_k

Thus v_k can be written as a linear combination of the other n-1 vectors (since t_k != 0).

Now suppose one of the vectors in the set v_i, denoted v_k, is a linear combination of the other n-1 vectors, so:

v_k = $\sum_{i=1}^{k-1}$ s_iv_i + $\sum_{i=k+1}^n$ s_iv_i

for some s ∈ Rⁿ, where s_k = 0. Then if we define t ∈ Rⁿ s.t. t_i = s_i for all values except t_k, which equals -1, then:

$\sum_{i=1}^n$ t_iv_i =

$\sum_{i=1}^{k-1}$ s_iv_i + (-1)v_k + $\sum_{i=k+1}^n$ s_iv_i =

$\sum_{i=1}^{k-1}$ s_iv_i + $\sum_{i=k+1}^n$ s_iv_i - ($\sum_{i=1}^{k-1}$ s_iv_i + $\sum_{i=k+1}^n$ s_iv_i) = 0

So $\sum_{i=1}^n$ t_iv_i = 0 when t != 0, so the set is linearly dependent.