I'm pretty sure trivial valuation over a field cannot be extended to a nontrivial one in a bigger field. Is there a simple way to show this without using the sledge hammer theorem on valuation extension over complete valued field?
Sure it can. Consider the xadic valuation on the field of Laurent series $k((x))$ over a field $k$, extending the trivial absolute value on $k$. 


Maybe you meant for the extension $L/K$ to be algebraic, in which case it is true that any extension of the trivial valuation on $K$ to $L$ is trivial. This clearly reduces to the case of a finite extension, and then  since a trivially valued field is complete  this follows from the uniqueness of the extended valuation in a finite extension of a complete field. Maybe you view this as part of the sledgehammer, but it's not really the heavy part: see e.g. p. 16 of http://math.uga.edu/~pete/8410Chapter2.pdf for the proof. (These notes then spend several more pages establishing the existence part of the result.) Addendum: Conversely, if $L/K$ is transcendental, then there exists a nontrivial extension on $L$ which is trivial on $K$. Indeed, let $t$ be an element of $L$ which is transcendental over $K$, and extend the trivial valuation to $K(t)$ by taking $v_{\infty}(P/Q) = deg(Q)  deg(P)$. (The completion of $K$ with respect to $v_{\infty}$ is the Laurent series field $K((t))$, so this is really the same construction as in Cam's answer.) Then I prove* in the same set of notes that any nonArchimedean valuation can be extended to an arbitrary field extension, so $v$ extends all the way to $L$ and is certainly nontrivial there, being already nontrivial on $K(t)$. *: not for the first time, of course, though I had a hard time finding exactly this result in the texts I was using to prepare my course. (This does use the sledgehammer.) 

