How to prove this formula for Lie derivative for differential forms The professor gave this formula without providing a proof. I would like to know how this can be derived. 

Let $X$ be a vector field, $w$ be a $p$-form. Then,
  $$L_X w(v_1,v_2,\ldots,v_p)=X(w(v_1,v_2,\ldots,v_p))-\sum_{i=1}^p w(v_1,\ldots,L_Xv_i,\ldots,v_p).$$

The definition for the Lie derivative is given by
$$L_Xw = \left.{{d}\over {dt}}\right|_{t=0} \phi_t^*w$$
where $\phi_t$ is the one parameter diffeomorphism group generated by $X$, and $\phi_t^*$ denotes the pull back. 
Thank you guys in advance for any answers and hints.
 A: The definition that you use for the Lie derivative, and the result you wish to deduce, both hold for any contravariant tensor field, so I will address the question for this more general situation.
On page $321$ of Lee's Introduction to Smooth Manifolds (second edition), he defines the Lie derivative of a covariant tensor as you have done. On the very next page he has Proposition $12.32\ (d)$ which states (I'm paraphrasing):

Let $A$ be a smooth contravariant $k$-tensor field on a smooth manifold $M$, and let $V, X_1, \dots, X_k$ be smooth vector fields on $M$. Then
$$\mathcal{L}_V(A(X_1, \dots, X_k)) = (\mathcal{L}_VA)(X_1, \dots, X_k) + \sum_{i=1}^kA(X_1, \dots, X_{i-1}, \mathcal{L}_VX_i, X_{i+1}, \dots, X_k).$$

In Corollary $12.33$, Lee points out that this formula can be rewritten as (again, I'm paraphrasing)
$$(\mathcal{L}_VA)(X_1, \dots, X_k) = V(A(X_1, \dots, X_k)) - \sum_{i=1}^kA(X_1, \dots, X_{i-1}, [V, X_i], X_{i+1}, \dots, X_k).$$
This follows immediately once you know $\mathcal{L}_Vf = Vf$ for any smooth function $f$ on $M$ (this is Proposition $12.32\ (a)$), and $\mathcal{L}_VX = [V, X]$ for any vector field $X$ on $M$ (this is Theorem $9.38$).
