This question can only have a subjective answer (which is actually fun, from time to time!), so here are a few personal remarks.
1) You are a dynamic PhD student working in 2012 under the supervision of Demailly, a world leader in complex algebraic geometry.
You have at your disposal a technology that didn't exist on Lefschetz' time: singular and De Rham cohomology, higher homotopy groups, Kähler manifolds, Hodge theory,...
Even the abstract notion of a finite-dimensional vector space had not been axiomatized.
So when you claim that the theorem is not that hard, you should not lose sight of the historic context in which Lefschetz "proved" his theorem in 1924.
2) I wrote "proved" in quotes, since as Sabbah diplomaticallty puts it, Lefschetz' proof was "insufficient".
So the theorem was not easy, even for Lefschetz.
3) The theorem has fascinated many Fields medalists and other giants who gave proofs of some version of the theorem: Andreotti, Frankel, Thom, Bott, Kodaira, Spencer, Artin, Grothendieck, Deligne.
This is certainly an indication of the depth of the theorem...
4) Like you I am enthusiastic about complex algebraic manifolds and am grateful for the transcendental methods , like Kähler theory, which allow us to study them.
However algebraic geometers also want to consider algebraic varieties in characteristic $p$, and there these transcendental tools unfortunately completely break down.
Hard Lefschetz for smooth varieties over finite fields was proved by Deligne only in 1980, after much preliminary work by himself and Grothendieck (cf. SGA7).
I would surmise that the terminology "Lefschetz vache" introduced by Grothendieck is to be understood in that context.
5) Finally even in the complex case, I find the proof of hard Lefschetz starting from scratch not so easy.
I'll let you and the other users judge by linking to a free online course of Sabbah on Hodge theory and hard Lefschetz (in the Introduction of which he writes the diplomatic remark mentioned above!)
Edit
Since this is a good-humoured, non-technical answer, I'll take the liberty of quoting the following picturesque metaphor by Lefschetz:
It was my lot to plant the harpoon of algebraic topology into the body of the whale of algebraic geometry
Let $\omega$ and $\omega'$ be two Kahler metrics on a compact Kahler manifold $X$. We denote by $L, *, \Lambda$ and $L', *', \Lambda'$ the operators in cohomology that the harmonic forms with respect to these metrics induce, and write $\omega$ and $\omega'$ for the Kahler classes in abuse of notation. (We won't speak of the actual metrics again.) It is obvious that if $\omega = \omega'$, then $L = L'$.
Claim. If $\omega = \omega'$, then $\operatorname{Ker} \Lambda = \operatorname{Ker} \Lambda'$.
It's enough to prove this for classes of a fixed degree $k$. Recall that for a class of degree $k \leq n$, we have
$$
[L^{r+1}, \Lambda] u = r(n-k+1-r) L^r u.
$$
Picking $r = n-k+1$ we get
$$
L^{n-k+1} \Lambda u = \Lambda L^{n-k+1} u.
$$
Recall that $\Lambda$ is injective on classes of degree $> n$, because $L$ is surjective on those classes and $\Lambda$ is its adjoint; and recall that $L^{n-k+1}$ is an isomorphism on classes of degree $k-1$. By comparing degrees on the left- and right-hand side above, we see that $\Lambda u = 0$ if and only if $L^{n-k+1}u = 0$. But if $\omega = \omega'$, then $L = L'$, so we conclude the claim.
Now consider the decompositions
$$
u = \sum L^j u_{k-j} = \sum (L')^j u_{k-j}'
$$
into products of primitive classes, where $u_{k-j}$ is primitive with respect to $\omega$, and $u_{k-j}'$ is primitive with respect to $\omega'$. By the claim, each $u_{k-j}'$ is actually primitive with respect to $\omega$, and by the unicity of the factors in the decomposition we conclude that $u_{k-j}' = u_{k-j}$ if $\omega = \omega'$. The formula for $*(L^k u_{k-j})$ then shows that if $\omega = \omega'$, then $* = *'$. As $\Lambda = *^{-1}L*$, we then conclude that $\Lambda' = \Lambda$ if $\omega = \omega'$.
Best Answer
For the completeness I write down the proof in detail.
The idea is for any real manifold, we can define the harmonic operator and the space of Harmonic form isomorphic to the deRham group with real coefficient. therefore the problem can be reduced to the problem of harmonic form.
For the Kahler setting we have the commutative relation $[L,\Delta_d]$ holds on the space $\mathcal{A}^{\cdot}_{\Bbb{C}}(X)$. However the key observation is that both $L$ and $d$ are real operator, that is if we restrict to the form with real coefficient $\mathcal{A}_{\Bbb{R}}^\cdot$ the image is still real form. $\Delta_d$ is real operator since $d$ is real operator ,$L$ is real operator since Kahler form is real (1,1) form.
Therefore we have the Kahler identity restrict to the real case. Since $L$ respect $\Delta_d$ (with real coefficient also complex coefficient), we have the isomorphism on the level of differential form $L^{n-k}: \mathcal{A}^k(X, \mathbb{R}) \cong \mathcal{A}^{2n-k}(X, \mathbb{R}) $ induce the isomorphism on the level of harmonic form.
Since harmonic form isomorphic to deRham group, we have the isomorphism on the deRham cohomology with real coefficient.
For the second statement, needs a little lemma: if $\alpha = \sum_i L^i \alpha_i$ being the decomposition such that $\alpha$ is harmonic form, then $\alpha_i$ is also harmonic form. This due to the fact that $\Delta_d$ commute with $L$, and the uniqueness of the decomposition.
For the third statement(which is called Lefschetz decomposition is compatible with the Hodge decomposition)
given $\alpha$ which is premitive k form, it has Hodge decomposition as $\alpha = \sum_{p+q = k} \alpha^{p,q}$ then apply the $\Lambda \alpha = 0$ then it will decomposition by bidegree, as $\Lambda$ is pure type (-1,-1) therefore all $\alpha^{p,q} \in \ker \Lambda$
(Here we use a little fact that kernel of complexified map is the kernel of original map tensor with $\Bbb{C}$. You can find the result on Conrad's note Theorem 2.9. here used the fact that the free module are flat.)