The coefficient of `x^(n)` in the expansion of `e^(x)` is

Options

A. `underset(r=n)overset(infty)Sigma (r^(n))/(r!)`
B. `(1)/(n!)underset(r=1)overset(infty)Sigma (r^(n))/(r!)`
C. `(1)/(n)underset(r=1)overset(infty)Sigma (r^(n))/(r!)`
D. none of these

Explanation

Let `e^(x)=z` Then
`e^(e^(x))=e^(Z)=underset(k=0)overset(infty)Sigma (z^(k))/(k!)=underset(k=0)overset(infty)Sigma(e^(x^(k)))/(k!)=underset(k=0)overset(infty)Sigma (e^(kx))/(k!)`
`rarr e^(e^(x))=(1+(e^(x)))/(1!)+(e^(2x))/(2!)+(e^(3x))/(3!)+…infty`
`rarr e^(e^(x))=1+(1)/(1!)+underset(n=0)overset(infty)Sigma(x^(n))/(n)+(1)/(2!) underset(n=0)overset(1)Sigma underset(n=0)overset(infty)Sigma (2x)^(n)/(n!)`
`+(1)/(3!)underset(n=0)overset(infty)Sigma(3x)^(n)/(n!)+...infty`
`therefore` coefficient of `x^(n) in e^(e^(x))`
`=(1)/(1!)(1)/(n!)+(1)/(2!)(2^(n))/(n!)+(1)/(3!)+...infty`
`(=(1)/(ni)(1)/(1!)+(2^(n))/(2!)+(3^(n))/(3!)+........infty)`
`=(1)/(ni)underset(r=1)overset(infty)Sigma(r^(n))/(r!)`