How would you account for the increasing oxidizing power in the series $VO_2^ + < MnO_4^ - $ ?

Hint:
Oxidizing power is defined by the ability of the compound to gain electrons; it can be compared by writing the electronic configuration of the compound and hence observing which compound can gain an electron to form a fully filled or half filled orbital to gain exceptional stability.

Complete step by step solution:
It is the variable oxidation states of transition elements that distinguish them from s and p block elements. S and p block elements have oxidation states either equal to their group number (G) or equal to (8-G). The transition elements on the other hand exhibit variable oxidation states.
This property is due to the fact that energy levels of 4d, 3d and 5d orbital are fairly close to those of 4s, 5s and 6s orbitals respectively. The relative stabilities of various oxidation states of 3d series elements can be correlated with the extra stability of $3{d^0}3{d^5}$ and $3{d^{10}}$ configurations due to completely paired electrons half filled and fully filled 3d orbital. Now let’s analyze the above given compounds and determine the oxidation state of both manganese and vanadium in their respective compounds.
Manganese has the electronic configuration of $3{d^5}4{s^1}4{p^1}$ in $MnO_4^ - $ and hence manganese has half filled orbital and is extremely stable whereas vanadium has the electronic configuration of $[Ar]3{d^3}4{s^2}$ in dioxovanadium.

 Manganese can exhibit the oxidation states from +2 to +7 and +7 oxidation state is its maximum oxidation state. Hence it is quite stable. The higher oxidation states of 4d and 5d series elements are generally more stable than those of the elements of 3d series which is why permanganate has more oxidizing power than dioxovanadium .the oxidation state is calculated by equating the charge on the compound to the sum of charges on its atoms multiplied by the number of atoms.
The above compounds are oxyanions formed by transition elements in their highest oxidation state with the more electronegative elements. Hence the above reasons explain why permanganate has higher oxidizing power than dioxovanadium.

Note:
Going down a sub group the stability of higher oxidation states increases while that of lower oxidation state decreases.