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`underset(r=n)overset(infty)Sigma (r^(n))/(r!)``(1)/(n!)underset(r=1)overset(infty)Sigma (r^(n))/(r!)``(1)/(n)underset(r=1)overset(infty)Sigma (r^(n))/(r!)`none of these

Solution :

Let `e^(x)=z` Then <br> `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!)` <br> `rarr e^(e^(x))=(1+(e^(x)))/(1!)+(e^(2x))/(2!)+(e^(3x))/(3!)+…infty` <br> `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!)` <br> `+(1)/(3!)underset(n=0)overset(infty)Sigma(3x)^(n)/(n!)+...infty` <br> `therefore` coefficient of `x^(n) in e^(e^(x))` <br> `=(1)/(1!)(1)/(n!)+(1)/(2!)(2^(n))/(n!)+(1)/(3!)+...infty` <br> `(=(1)/(ni)(1)/(1!)+(2^(n))/(2!)+(3^(n))/(3!)+........infty)` <br> `=(1)/(ni)underset(r=1)overset(infty)Sigma(r^(n))/(r!)`**Introduction**

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