Yes. By definition. \( \vartheta_n(n) = 1 \) and \( \vartheta_k(n) = 0 \,\,\Leftrightarrow\,\, k \neq n\,\,k \in \mathbb{N}_0 \) therefore the rest of the product disappears at natural values. That's the whole trick 
. I'd take the limit as \( \lim_{h \to n}\vartheta_n (h) = \vartheta_n(n) = 1 \) just for technicality sake.
\( x\,\,\bigtriangleup_0\,\, y = (x + y)^1 \cdot (x \cdot y)^0 \cdot (x^y)^0 \cdot ... \)
\( x \,\,\bigtriangleup_1\,\, y = (x + y)^0 \cdot (x \cdot y)^1 \cdot (x^y)^0 \cdot ... \)
It's real for real \( s \) and complex for complex \( s \) and natural for natural \( s \). Very nice.
. I'd take the limit as \( \lim_{h \to n}\vartheta_n (h) = \vartheta_n(n) = 1 \) just for technicality sake.\( x\,\,\bigtriangleup_0\,\, y = (x + y)^1 \cdot (x \cdot y)^0 \cdot (x^y)^0 \cdot ... \)
\( x \,\,\bigtriangleup_1\,\, y = (x + y)^0 \cdot (x \cdot y)^1 \cdot (x^y)^0 \cdot ... \)
It's real for real \( s \) and complex for complex \( s \) and natural for natural \( s \). Very nice.
