Equivalent fomulations of Bott periodicity Is there an easy way to see the equivalence of the two statements of Bott periodicity.
$$BU \times \mathbb{Z} \simeq \Omega^2BU$$ and 
$$K(X)\otimes K(S^2) \cong K(X\times S^2)$$ 
 A: The easy way for me to think about this things is via the Yoneda lemma. This works better with reduced $K$-theory. This is defined for a pointed space $X$ as
$$ \tilde K^0 (X) := ker(K^0(X)\to K^0(*))$$
It is ``easy'' to see that your second statement reduces to the fact that multiplication by the generator of $\tilde K^0(S^2)$ induces an isomorphism $\tilde K^0(X) \cong \tilde K^0(X\wedge S^2)$
(this can be seen in many ways, one is noting that the projections of a product to two factors have canonical pointed sections and so induce direct summands in $K$-theory).
The key fact about reduced $K$-theory is that it is representable. That is there is a natural isomorphism
$$ \tilde K^0(X) \cong [X,BU\times \mathbb{Z}]_*$$
where $[,]_*$ are the homotopy classes of maps. So your second statement says that there is a natural isomorphism
$$[X,BU\times \mathbb{Z}]_* \cong [\Sigma^2 X,BU\times\mathbb{Z}]_* $$
By using the suspension-loopspace adjunction you get a natural isomorphism
$$ [X, BU\times \mathbb{Z}]_* \cong [X,\Omega^2(BU\times\mathbb{Z})]_*$$
Finally the Yoneda lemma tells us that this must correspond to an homotopy equivelence $BU\times \mathbb{Z}\sim \Omega^2(BU\times\mathbb{Z})$, which is your first statement.
If you look closely all steps are reversible, so the two statements are in fact equivalent.
EDIT: For future reference I am putting the steps to recover the unreduced statement from the reduced statement
I am adding the steps on how to prove the reduced formulas and the unreduced formulas to be equivalent. First note that the basepoint inclusion $*\to X$ has a retraction $X\to *$, so
$$ K(X) = \tilde K(X)\oplus \mathbb{Z}$$
Moreover, if $X,Y$ are pointed spaces the inclusion $X=X\times * \to X\times Y$ has a retraction, so the induced map on $K$-theory
$$ \tilde K^0(X\times Y) \to \tilde K^0(X) $$
has a section. Since the same is true for $Y$ from the additivity of $K$-theory it follows that the map induced by the inclusion $X\vee Y\to X\times Y$
$$ \tilde K^0(X\times Y)\to \tilde K^0(X\vee Y) =  \tilde K^0(X)\oplus \tilde K^0(Y)$$
splits. Now consider the cofiber sequence
$$ X\vee Y \to X\times Y \to X\wedge Y$$
This induces an exact sequence
$$ \tilde K^0(\Sigma(X\times Y)) \to \tilde K^0(\Sigma X\vee \Sigma Y) \to \tilde K^0(X\wedge Y) \to \tilde K^0(X\times Y) \to \tilde K^0(X\vee Y)$$
We now that the rightmost arrow splits and the rightmost arrow splits too by a similar argument (in fact it splits even at the space level). So we have the desired isomorphism
$$ K(X\times Y) = \mathbb{Z}\oplus\tilde K^0(X\times Y) = \mathbb{Z}\oplus\tilde K^0(X) \oplus \tilde K^0(Y) \oplus \tilde K^0(X\wedge Y) $$
