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function as an infinite ‘product. 
nk n = -2(- l) h 
[h = 0, 1, 2, 3, ... (p. — 1)] 
to p terms, because there are as many divisors 8 h as there are 
combinations of p things li at a time. Hence 
nk n = — (1 — l)'* + (— l)' 4 = (- 1)^, 
in accordance with formula (ii) : and since, when p = l, we have 
k n = — l/n, this case is proved in general. 
To prove case (iii), suppose first that m = 1, and write n — 2P. 
Then by (2) 
nk n = X8k s — X2&ko S 4- Pkp, 
where the sums apply to all the proper divisors of P. Assuming 
(ii) and (iii) to hold good for the right-hand side, 
nk n = Pk P = (- l)* 4 , 
because, by hypothesis, k s = 2 k 2S . 
Next suppose ni = 2, and write n = 4 P. Then, with the same 
notation, we obtain from (2) 
nk n = %8k s + ^28k. 2S — 24 8k 4 $ -1- Pkp 4- 2 Pk. 2 p 
= (- ly- 1 + (- iy~' - 2 (- iy~' + (-iy + (-iy 
= 2 (— ly. 
Similarly, for m — 3, 
nk n = %8k s 4- 2 2 8k 2S 4 %48k 4& — 28 8k sS 
4" Pkp 4 2P k 2 p 4- 4 P k 4 p 
= (1 +1 + 2-4) (— l)' 4-1 4- (1 4- 1 + 2) (-1)* 
= 4 (— l)' 4 , 
and it is now easy to see that formula (iii) is correct, because when 
n = 2 p, 4 p respectively, p being an odd prime, formula (2) gives 
results in accordance with (iii). 
To prove (iv), put 
n=p h m (li> 1) 
where p is an odd prime, and m is prime to p. Let 2 a P be the 
greatest factor of m which has no odd square divisor. Then 
assuming the truth of (iv) for all integers less than n, equation (2) 
may be written 
nk n = 2 (- 1 ) n/s 8k s + 2 (- 1 ) nlpS p8k pS> 
where the summation extends to all divisors of 2 a P, including this 
number itself. Now 
(_ iyilps _ i WS } 
because p is odd ; and by formulas (ii) and (iii) 
p8k pS = — 8k$: 
nk n = 2 (— l) nls {S&6 4- p8k pS \ = 0. 
hence 
