Show that a polynomial $f(x)$ over a field $k$ is irreducible if the polynomial $f(x+1)$ is irreducible I was thinking of using contradiction by assuming that that $f(x)$ is reducible but I really don't know how to continue from that idea.
Assume that $f(x)$ is reducible. Then:

$\exists \ h(x), \ g(x) \in \mathbb K[x]$ such that $f(x) = g(x)h(x)$.

Then: $f(x + 1) = g(x+1)h(x+1)$, and so reducible. A contradiction.
Let's prove the contrapositive:  That if $f(x)$ is reducible then $g(x) \equiv f(x+1)$ is also reducible.  That is logically equivalent to the statement to be proven.

Since $f(x)$ is reducible there exists some $x_0\in K:f(x_0)=0$.  Then since $1 \in K$ and $K$ is a field, $y\equiv x_0-1 \in K$.  And 
$$
g(y) = f((x_0-1)+1) = f(x_0) = 0
$$ so $g(x)$ is reducible.

EDITED in response to Bill's comment:

This "proof" is invalid.  Yes, if $f(x)$ has a zero then $g(x)$ is irreducible.  But we don't know that $f(x)$ has a zero; all we know is that it is irreducible.  So "step 1.5" in the above proof is hogwash.  Sorry.