(G.M. 1/2012)

Mai intai impartim ambele parti cu n! .
Deoarece stim cum cresc si apoi descresc coeficientii binomiali,
ajunge sa ne legam de cazurile:

n = 2N si k=N si
n = 2N+1 si k=N (sau k=N+1, acelasi lucru).

De aici incolo terminam repede inductiv (dupa N in fiecare caz) sau cu Stirling.

De exemplu:

? for( n=1,9, print( "(1+1/",n,")^",n," ~ ", (1.+1./n)^n) )
(1+1/1)^1 ~ 2.000000000000000000000000000
(1+1/2)^2 ~ 2.250000000000000000000000000
(1+1/3)^3 ~ 2.370370370370370370370370370
(1+1/4)^4 ~ 2.441406250000000000000000000
(1+1/5)^5 ~ 2.488320000000000000000000000
(1+1/6)^6 ~ 2.521626371742112482853223594
(1+1/7)^7 ~ 2.546499697040713113947905574
(1+1/8)^8 ~ 2.565784513950347900390625000
(1+1/9)^9 ~ 2.581174791713197181990031508