Since $4(n^2+n+41)=(2n+1)^2+163$, one way would be to methodically show that $-163$ is a quadratic *non*-residue mod all (odd) primes less than $41$.

**Added later**: The OP has asked for further details, so here goes.

If $p\mid n^2+n+41$, then $(2n+1)^2+163\equiv0$ mod $p$, which means that $-163$ is a square mod $p$. Now the values of $n^2+n+41$ for $n\lt40$ are all less than $40^2+40+41=1681$, so if any of them were composite, it would have to have a prime factor less than $\sqrt{1681}=41$. That prime factor would have to be odd, since it's clear that $n^2+n+41$ is always odd. This would require $-163$ to be a quadratic residue for some odd prime less than $41$. So it suffices to compute the Legendre symbol $({-163\over p})$ for $p=3,5,7,11,13,17,19,23,29,31$, and $37$, and show that it's $-1$ in each case.

This is a bit tedious, but less so than directly checking the primality of the $39$ numbers $(1^1+1+41), (2^2+2+41),\ldots,(39^2+39+41)$. For example,

$$\left({-163\over17}\right)=\left({7\over17}\right)=\left({17\over7}\right)=\left({3\over7}\right)=-\left({7\over3}\right)=-\left({1\over3}\right)=-1$$

where the string of equalities here makes use (twice) of the Law of Quadratic Reciprocity.