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}}.</ref> Specifically, define the sequence <math> s_{n}</math> for all <math>n \ge 0</math> as follows: <math>s_{n} = \prod_{k = 0}^{n} {n \choose k} = \prod_{k = 0}^{n} \frac{n!}{k!(n-k)!}</math> {{pb}} Then, the ratio of successive row products is <math display="block">\frac{s_{n+1}}{s_{n}} =
\frac{ \displaystyle (n+1)!^{n+2} \prod_{k = 0}^{n + 1} \frac{1}{k!^2}}{\displaystyle n!^{n+1}\prod_{k=0}^{n}{\frac{1}{k!^2}}} = \frac{(n + 1)^n}{n!}</math> and the ratio of these ratios is <math display="block">\frac{s_{n + 1} \cdot s_{n - 1}}{s_{n}^{2}} =
\left( \frac{n + 1}{n} \right)^n, ~ n\ge 1.</math> The right-hand side of the above equation takes the form of the limit definition of [[e (mathematical constant)|<math>e</math>]] <math display="block">e =\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{n}.</math>
* [[pi|<math>\pi</math>]] can be found in Pascal's triangle by use of the [[Nilakantha Somayaji|Nilakantha]] infinite series.<ref>{{citation
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