Graph Theory Fun Problem Show that in any graph $G$ with min-degree $k$ ($k \geq 1$ duh!) you can find as its subgraph any tree on $k+1$ vertices.

I have not been able to solve the question so far. However, I would like if someone can invoke the probabilistic method to prove the above statement (as in if someone can show that the probability of not finding some tree on $k+1$ vertices as some subgraph of $G$ is less than $1$. My earlier efforts aimed at trying to show that we can somehow show that this graph contains $K_{k+1}$ which was obviously false. Next, I tried showing that a sequence of edge contractions can give a $K_{k+1}$. Though I have not shown it so far, I was wondering what good that result might have been. So far, sadly, I have gotten nowhere :(

Sorry, if the question is too dumb.
)
Let $G$ be the host graph and $T$ the tree. Try "growing" $T$ in $G$. At each step, mark in $G$ the subtree $T'$ of $T$ that has been grown so far. Using the bounds on $\delta(G)$ and $\vert V(T)\vert$, argue that we can always expand the marked subgraph if it is not yet isomorphic to $T$. 

Oh, but you wanted the Probabilistic Method. Here we go: Toss a coin $k$ times, throw out the results, and then do the above (old joke).