Editing Transcendental number (section)

====Lemma 1====
''There are arbitrarily large {{mvar|k}} such that <math>\ \tfrac{P}{k!}\ </math> is a non-zero integer.''

'''Proof.''' Recall the standard integral (case of the [[Gamma function]])
<math display=block>
  \int^{\infty}_{0} t^{j} e^{-t} \,\mathrm{d}t = j!
</math>
valid for any [[natural number]] <math>j</math>. More generally,

: if <math> g(t) = \sum_{j=0}^m b_j t^j </math> then <math> \int^{\infty}_{0} g(t) e^{-t} \,\mathrm{d}t = \sum_{j=0}^m b_j j! </math>.

This would allow us to compute <math>P</math> exactly, because any term of <math>P</math> can be rewritten as
<math display=block>
  c_{a} e^{a} \int^{\infty}_{a} f_k(x) e^{-x} \,\mathrm{d}x =
  c_{a} \int^{\infty}_{a} f_k(x) e^{-(x-a)} \,\mathrm{d}x =
  \left\{ \begin{aligned} t &= x-a \\ x &= t+a \\ \mathrm{d}x &= \mathrm{d}t \end{aligned} \right\} =
  c_a \int_0^\infty f_k(t+a) e^{-t} \,\mathrm{d}t
</math>
through a [[Integration by substitution|change of variables]]. Hence
<math display="block">
  P = \sum_{a=0}^n c_a \int_0^\infty f_k(t+a) e^{-t} \,\mathrm{d}t
  = \int_0^\infty \biggl( \sum_{a=0}^n c_a f_k(t+a) \biggr) e^{-t} \,\mathrm{d}t
</math>
That latter sum is a polynomial in <math>t</math> with integer coefficients, i.e., it is a linear combination of powers <math>t^j</math> with integer coefficients. Hence the number <math>P</math> is a linear combination (with those same integer coefficients) of factorials <math>j!</math>; in particular <math>P</math> is an integer.

Smaller factorials divide larger factorials, so the smallest <math>j!</math> occurring in that linear combination will also divide the whole of <math>P</math>. We get that <math>j!</math> from the lowest power <math>t^j</math> term appearing with a nonzero coefficient in <math>\textstyle \sum_{a=0}^n c_a f_k(t+a) </math>, but this smallest exponent <math>j</math> is also the [[Multiplicity (mathematics)#Multiplicity of a root of a polynomial|multiplicity]] of <math>t=0</math> as a root of this polynomial. <math>f_k(x)</math> is chosen to have multiplicity <math>k</math> of the root <math>x=0</math> and multiplicity <math>k+1</math> of the roots <math>x=a</math> for <math>a=1,\dots,n</math>, so that smallest exponent is <math>t^k</math> for <math>f_k(t)</math> and <math>t^{k+1}</math> for <math>f_k(t+a)</math> with <math> a>0 </math>. Therefore <math>k!</math> divides <math>P</math>.

To establish the last claim in the lemma, that <math>P</math> is nonzero, it is sufficient to prove that <math>k+1</math> does not divide <math>P</math>. To that end, let <math>k+1</math> be any [[prime number|prime]] larger than <math>n</math> and <math>|c_0|</math>. We know from the above that <math>(k+1)!</math> divides each of <math> \textstyle c_a \int_0^\infty f_k(t+a) e^{-t} \,\mathrm{d}t </math> for <math> 1 \leqslant a \leqslant n </math>, so in particular all of those ''are'' divisible by <math>k+1</math>. It comes down to the first term <math> \textstyle c_0 \int_0^\infty f_k(t) e^{-t} \,\mathrm{d}t </math>. We have (see [[falling and rising factorials]])
<math display=block>
  f_k(t) = t^k \bigl[ (t-1) \cdots (t-n) \bigr]^{k+1} =
  \bigl[ (-1)^{n}(n!) \bigr]^{k+1} t^k + \text{higher degree terms}
</math>
and those higher degree terms all give rise to factorials <math>(k+1)!</math> or larger. Hence
<math display=block>
  P \equiv 
  c_0 \int_0^\infty f_k(t) e^{-t} \,\mathrm{d}t \equiv
  c_0 \bigl[ (-1)^{n}(n!) \bigr]^{k+1} k! \pmod{(k+1)}
</math>
That right hand side is a product of nonzero integer factors less than the prime <math>k+1</math>, therefore that product is not divisible by <math>k+1</math>, and the same holds for <math>P</math>; in particular <math>P</math> cannot be zero.