Problem: Prove that for every nonnegative integer $n$,
the number $7^{7^n}+1$ is the product
of at least $2n+3$ (not necessarily distinct) primes.

Solution: We prove this by induction on $n$ by showing that
\[ \frac{X^7+1}{X+1} = X^6 - X^5 + \dots + 1 \]
is never prime for $X = 7^{7^n}$,
hence we gain at least two additional prime factors
whenever we increase $n$ by one.

Indeed, the quotient may be written as
\[ \left( X+1 \right)^6 - 7X \cdot (X^2+X+1)^2 \]
which becomes a difference of squares, hence composite.