Prove that if $E[X_1^p]<\infty$, then $\frac{\max_{1\le i\le n} X_i}{n^{1/p}} \rightarrow 0$ in probability where $\{X_n\}$ is i.i.d and non-negative Suppose $\{X_n\}_{n\geq 1}$ are iid and non negative. Define $M_n=\max \limits_{i=1,\ldots,n}\{X_i\}.$ 
Prove if $E[X_1^p]<\infty$, then $\frac{M_n}{n^{1/p}}\rightarrow 0$ in probability.
As a previous result I have that $P[M_n>x]\leq nP[X_1>x].$ Is this result useful to prove this convergence? I would appreciate any hint.

You can use the result $P[M_n>x]\leq nP[X_1>x]$. Observe that
$$P[M_n/n^{1/p}>\epsilon]=P[M_n>\epsilon n^{1/p}]
\le nP[X>\epsilon n^{1/p}]=nP[X^p>\epsilon^pn].$$ Now use the inequality $P[Y>a]\le E\left[\frac YaI(Y>a)\right]$ to conclude $$nP[X^p>\epsilon^pn]\le E\left[\frac{X^p}{\epsilon^p}I(X^p>\epsilon^pn)\right],$$ and finally argue that the RHS tends to $0$ as $n\to\infty$.
