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@Lopsy
[==  
We have
f\(n\) = \\sum\_{k=0}\^N m\_k n\^k.
Therefore
f\(f\(n\)+1\)/f\(n\) = 1/f\(n\) \* \\sum\_{k=0}\^N m\_k \(f\(n\)+1\)\^k = \\sum\_{k=0}\^N \\sum\_{l=0}\^k \(k choose l\) m\_k f\(n\)\^{l-1},
by Newton's formula.

Change the order of summation to get
f\(f\(n\)+1\)/f\(n\) = \\sum\_{l=0}\^N \\sum\_{k=l}\^N \(...\)
and then split the sum over l into the l=0 term and the l\>0 terms. The latter sum to an integer as each term is individually an integer, so we only need to check the l=0 term.

This is just
\\sum\_{k=0}\^N m\_k/f\(n\) = f\(1\)/f\(n\).

By hypothesis, f\(n\) is strictly larger than f\(1\) \(in which case f\(1\)/f\(n\) is not an integer and f\(n\) does not divide f\(f\(n\)+1\)\) unless n = 1 or f is a zero-degree polynomial \(a constant\).
==]

@Lopsy
[==We have
f(n) = \sum_{k=0}^N m_k n^k.
Therefore
f(f(n)+1)/f(n) = 1/f(n) * \sum_{k=0}^N m_k (f(n)+1)^k = \sum_{k=0}^N \sum_{l=0}^k (k choose l) m_k f(n)^{l-1},
by Newton's formula.

Change the order of summation to get
f(f(n)+1)/f(n) = \sum_{l=0}^N \sum_{k=l}^N (...)
and then split the sum over l into the l=0 term and the l>0 terms. The latter sum to an integer as each term is individually an integer, so we only need to check the l=0 term.

This is just
\sum_{k=0}^N m_k/f(n) = f(1)/f(n).

By hypothesis, f(n) is strictly larger than f(1) (in which case f(n) does not divide f(f(n)+1)) unless n = 1 or f is a zero-degree polynomial (a constant).
==]

@Lopsy
[==We have
f(n) = \sum_{k=0}^N m_k n^k.
Therefore
f(f(n)+1)/f(n) = 1/f(n) * \sum_{k=0}^N m_k (f(n)+1)^k = \sum_{k=0}^N \sum_{l=0}^k (k choose l) m_k f(n)^{l-1},
by Newton's formula.

Change the order of summation to get
f(f(n)+1)/f(n) = \sum_{l=0}^N \sum_{k=l}^N (...)
and then split the sum over l into the l=0 term and the l>0 terms. The latter aresum to an integer, becaus each they arem is individumally ofan integers, so we only need to check the l=0 term.

This is just
\sum_{k=0}^N m_k/f(n) = f(1)/f(n).

By hypothesis, f(n) is strictly larger than f(1) unless n = 1 or f is a zero-degree polynomial (a constant).==]

@Lopsy
[==We have
f(n) = \sum_{k=0}^N m_k n^k.
Therefore
f(f(n)+1)/f(n) = 1/f(n) * \sum_{k=0}^N m_k (f(n)+1)^k = \sum_{k=0}^N \sum_{l=0}^k (k choose l) m_k f(n)^{l-1},
by Newton's formula.

Change the order of summation to get
f(f(n)+1)/f(n) = \sum_{l=0}^N \sum_{k=l}^N (...)
and then split the sum over l into the l=0 term and the l>0 terms. The latter are an integer, because they are sum of integers, so we only need to check the l=0 term.

This is just
\sum_{k=0}^N m_k/f(n) = f(1)/f(n).

By hypothesis, f(n) is strictly larger than f(1) unless n = 1 or f is a zero-degree polynomial (a constant).==]

@Lopsy
[==We have
f(n) = \sum_{k=0}^N m_k n^k.
Therefore
f(f(n)+1)/f(n) = 1/f(n) * \sum_{k=0}^N m_k (f(n)+1)^k = \sum_{k=0}^N \sum_{l=0}^k (k choose l) m_k f(n)^{l-1},
by Newton's formula. Change the order of summation to get
f(f(n)+1)/f(n) = \sum_{l=0}^N \sum_{k=l}^N (...)
and then split the sum over l into the l=0 term and the l>0 terms. The latter are an integer, because they are sum of integers, so we only need to check the l=0 term.
This is just
\sum_{k=0}^N m_k/f(n) = f(1)/f(n).
By hypothesis, f(n) is strictly larger than f(1) unless n = 1 or f is a zero-degree polynomial (a constant).==]

@Lopsy
We have
f(n) = \sum_{k=0}^N m_k n^k.
Therefore
f(f(n)+1)/f(n) = 1/f(n) * \sum_{k=0}^N m_k (f(n)+1)^k = \sum_{k=0}^N \sum_{l=0}^k (k choose l) m_k f(n)^{l-1},
by Newton's formula. Change the order of summation to get
f(f(n)+1)/f(n) = \sum_{l=0}^N \sum_{k=l}^N (...)
and then split the sum over l into the l=0 term and the l>0 terms. The latter are an integer, because they are sum of integers, so we only need to check the l=0 term.
This is just
\sum_{k=0}^N m_k/f(n) = f(1)/f(n).
By hypothesis, f(n) is strictly larger than f(1) unless n = 1 or f is a zero-degree polynomial (a constant).