Number of connected components after removing $t$ edges from K-edge-connected graph $k,s,t$ are positive integers such that $k(s+1)&gt;2t$.

prove: if $G$ k-edge-connected graph then removing any $t$ edges from $G$ will yield a graph with at most $s$ connected components.

question from old exam i'm trying to solve .... have no idea where to start . 
the graph is k-connected so i guess i'm expected to use "Whitney theorem" or "Manger theorem" but i don't see how i can derive something about number of connected components using mentioned theorems and the given inequality 

after some thought i think i need to use induction here .
some trail and error led me to this solution hope its a valid one . 

given $s$ number of connected components after the removal of $t$ edges .
each component had at least $k$ edges what were connecting vertices in it to the other vertices of the original graph $G$ , moreover each removed edge were connected to 2 vertices
so it was connecting 2 components before it was removed. from this we can derive the wanted inequality 

$$sk &lt; 2t &lt; sk + k$$

where:


$sk$ the minimal number of edges you need to remove to yield $s$ connected components
$2t$ the number of removed edges 
$sk + k = k(s+1)$ the number of edges you would need to remove if there were $s+1$ connected components
some trail and error led me to this solution hope its a valid one . 

given $s$ number of connected components after the removal of $t$ edges .
each component had at least $k$ edges what were connecting vertices in it to the other vertices of the original graph $G$ , moreover each removed edge were connected to 2 vertices
so it was connecting 2 components before it was removed. from this we can derive the wanted inequality 

$$sk &lt; 2t &lt; sk + k$$

where:


$sk$ the minimal number of edges you need to remove to yield $s$ connected components
$2t$ the number of removed edges 
$sk + k = k(s+1)$ the number of edges you would need to remove if there were $s+1$ connected components