Suppose there exists a graph with longest path of $m < k$ from $v_1, v_2,....,v_{m+1}.$
$v_{m+1}$ is the last vertex of the path that means there is no other vertex in the graph that is not already included in the path. This means if $v_{m+1}$ is connected to all the vertices of the path then too deg($v_{m+1}$) $< k$, this contradicts our assumption that each vertex having degree $k > m$.