Igor is essentially right. If $n > 2$ and the embedding is convex, then the second fundamental form exists almost everywhere. Using the Gauss equations per Robert's answer, it extends uniquely and smoothly to everywhere. The rest is straightforward. I'm omitting it because I'm typing this on an iPhone.

The $C^1$ but non-convex case is due to Nash (who did it in codimension 2) and Kuiper (who did it in codimension 1). It is not a special case of smooth Nash embedding theorem.