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Let $X$ be a random variable; then$$ E[\left\vert X\right\vert ^{k}]\leq M^{k}\text{ \ \ \ \ \ \ }\forall k\geq 1,k\in \mathbf{N} $$
if and only if,$\forall i\geq 0,i\in \mathbf{N}$ $$ E\left[ \left\vert X\right\vert ^{k}\right] \leq E\left[ \left\vert X\right\vert ^{i}\right] M^{k-i}\text{ \ \ \ \ \ \ \ }\forall k\geq i,k\in \mathbf{N} $$
Proof. a) $(E\left[ \left\vert X\right\vert ^{k}\right] \leq E\left[ \left\vert X\right\vert ^{i}\right] M^{k-i}$ $\Longrightarrow $ $% E[\left\vert X\right\vert ^{k}]\leq M^{k})$ It's enough to take $i=0$ and the thesis follows easily.
b) $(E[\left\vert X\right\vert ^{k}]\leq M^{k}$ $\Longrightarrow E\left[ \left\vert X\right\vert ^{k}\right] \leq E\left[ \left\vert X\right\vert ^{i}% \right] M^{k-i})$ Let $1\leq i\leq k$ (the case $i=0$ is trivial). Then, using Cauchy-Schwarz inequality $N$ times, one has:\begin{eqnarray*} E[\left\vert X\right\vert ^{k}] &=&E\left[ \left\vert X\right\vert ^{\frac{i% }{2}}\left\vert X\right\vert ^{k-\frac{i}{2}}\right] \\ &\leq &E\left[ \left\vert X\right\vert ^{i}\right] ^{\frac{1}{2}}E\left[ \left\vert X\right\vert ^{2k-i}\right] ^{\frac{1}{2}} \\ &=&E\left[ \left\vert X\right\vert ^{i}\right] ^{\frac{1}{2}}E\left[ \left\vert X\right\vert ^{\frac{i}{2}}\left\vert X\right\vert ^{2k-\frac{3}{2% }i}\right]
^{\frac{1}{2}} \\ &\leq &E\left[ \left\vert X\right\vert ^{i}\right] ^{\left( \frac{1}{2}+% \frac{1}{4}\right) }E\left[ \left\vert X\right\vert ^{4k-3i}\right] ^{\frac{1% }{4}} \\ &\leq &E\left[ \left\vert X\right\vert ^{i}\right] ^{\left( \frac{1}{2}+% \frac{1}{4}+\frac{1}{8}\right) }E\left[ \left\vert X\right\vert ^{\left( 8k-7i\right) }\right] ^{\frac{1}{8}} \\ &&... \\ &\leq &E\left[ \left\vert X\right\vert ^{i}\right] ^{\left( \sum_{m=1}^{N}% \frac{1}{2^{m}}\right) }E\left[ \left\vert X\right\vert ^{2^{N}k-\left( 2^{N}-1\right) i}\right] ^{\frac{1}{2^{N}}} \\ &=&E\left[ \left\vert X\right\vert ^{i}\right] ^{\left( 1-\frac{1}{2^{N}}% \right) }E\left[ \left\vert X\right\vert ^{2^{N}\left( k-i\right) +i}\right] ^{\frac{1}{2^{N}}} \\ &\leq &E\left[ \left\vert X\right\vert ^{i}\right] ^{\left( 1-\frac{1}{2^{N}}% \right) }M^{\left( k-i\right) +\frac{i}{2^{N}}}, \end{eqnarray*} and since this must hold for any $N$ we obtain
$$ E[\left\vert X\right\vert ^{k}]\leq E\left[ \left\vert X\right\vert ^{i}% \right] M^{k-i} $$ 
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