Proof using the binomial theorem

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The discussion focuses on using the binomial theorem to prove the equation (1 + 1/n)^n = 1 + sum(k=1 to n) [1/k! product(r=0 to k-1) (1 - r/n)] for positive integers n. An attempt to manipulate the left-hand side using binomial coefficients and products is made, but the contributor expresses frustration with the complexity of the expressions. They rewrite the product on the right-hand side to relate it to factorials but feel that the progress is unsatisfactory. Another participant suggests that the terms on the right correspond to the expanded terms of the left, indicating a potential path forward. The conversation highlights the challenges in proving the identity while exploring connections between the two sides of the equation.
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Homework Statement


Use the binomial theorem to rpove that for n a positive integer we have:

(1 + 1/n)^n = 1 + sum(k=1 to n) [1/k! product(r=0 to k-1) (1 - r/n)]

The Attempt at a Solution



(1 + 1/n)^n = 1 + sum(k=1 to n) (n choose r) 1^n-k (1/n)^k, where (n choose r) = n!/r!(n - r)!, the binomial coefficients.

I'm trying to fit "n!/r!(n - r)!" to an expression that involves the products, since product(k=1 to n) n = n!

The product on the RHS I rewrite it as: product(r=0 to k-1) [(n - r)/n]

[product(r=0 to k-1) (n - r)] x [product(r=0 to k-1) (1/n)]

=> (n - r)! (1/n)!

So...

(1 + 1/n)^n = 1 + sum(k=1 to n) [1/k! (n-r)! (1/n)!] or (1 + 1/n)^n = 1 + sum(k=1 to n) [(n-r)!/k!n!]

I don't like this. I feel that all of this is taking me nowhere.

Any ideas will be very appreciated.
 
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Hi

look at the snapshot i have attached. in the second equation, its clear that we can express
it as the term on RHS in your expression.
 

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Each of the terms of the RHS sum corresponds to one of the expanded terms of the LHS.
 

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