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We can reorder the summation as $\displaystyle\sum_{m=1}^{5} \sum_{n=1}^{5} m^{4n + 1}.$

Now for each of $1 \le m \le 5$ we find that $m^{4n + 1} \pmod{10} = m$ for all non-negative integers $n,$ and thus $\displaystyle\sum_{n=1}^{5} (m^{4n + 1} \pmod{10}) = 5m$ for each $m.$

Summing this over $m$ then gives us a result of

$5*\dfrac{5*(5 + 1)}{2} = 75 \equiv 5 \pmod{10},$ i.e., the units digit is $\boxed{5}.$

For n=1, we get 1, 2, 3, 4 ,5. Each of the digit from 1 to 5 is cyclic. 1, and 5 remain 1 and 5 for any power. 2 and 3 are cyclic with period 4, the same as $\color{#D61F06}{4}$ n. 4 has a period of 2, so it will be back also after 4 cycle. So for all n, for m=1 to 5, we will have 1, 2, 3, 4, 5. So sum = 5 * (1+5) * 5 /2=75. Last digit is 5.

how u solved i havent understood