Introducing a new operation

Complete the product: $a + ab + b = (a + 1)(b + 1) - 1$ . This suggests that $(a\star b) + 1 = (a + 1)(b+1),$ so that $(a\star b \star c \star \cdots \star z) + 1 = (a+1)(b+1)(c+1)\cdots (z+1),$ and $a^{\star n} + 1 = (a+1)^n.$ Thus $7^{\star 7} = (7 + 1)^7 - 1 = \boxed{8^7 - 1}.$

From the first few $n$ , we find that $a^{\star n} = (a+1)^n - 1$ . Let us prove by induction that the claim is true for all $n \ge 1$ .

Proof: For $n=1$ , $a^{\star 1} = (a+1)^1 - 1 = a$ as defined. Therefore, the claim is true for $n=1$ . Now assuming that the claim is true for $n$ , then we have:

$\begin{aligned} a^{\star (n+1)} & = a \star a^{\star n} \\ & = a + a(a^{\star n}) + a^{\star n} \\ & = a + a((a+1)^n - 1) + (a+1)^n - 1 \\ & = a + a(a+1)^n - a + (a+1)^n - 1 \\ & = (a+1)^{n+1}-1 \end{aligned}$ .

Therefore, the claim is also true for $n+1$ and hence for all $n \ge 1$ . Then we have $7^{\star 7} = (7+1)^7 - 1 = \boxed{2097151}$ .

We can see, with some counting, that the following formula holds (proof at the end):

$\displaystyle{a^{\star n}=\sum_{k=1}^{n} {{n}\choose{k}}a^k}$

In particular, if $a=1$ , we have

$\displaystyle{1^{\star n}=\sum_{k=1}^{n} {{n}\choose{k}}=2^n-1}$

And since $7=2^3-1=1^{\star 3}$ , we have:

$\displaystyle{7^{\star 7}=(1^{\star 3})^{\star 7}=1^{\star 21}=2^{21}-1=\boxed{2097151}}$

Note: Proof of the fact that $\left(1^{\star 3}\right)^{\star 7}=1^{\star 21}$ is left to the kind reader. Does this power property apply to every base or just to 1?

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Now, as to why the star power formula holds. We are just going to prove this by induction. It is in fact obvious that the formula holds for $n=2$ , as $\displaystyle{a^{\star2}=a\star a=a^2+2a= {{2}\choose{2}}a^2+{{2}\choose{1}}a}$

Let's suppose that it holds for $n$ and let's prove it for $n+1$ .

$\displaystyle{a^{\star (n+1)}=a^{\star n}\star a=\left(\sum_{k=1}^{n} {{n}\choose{k}}a^k\right)\star a=\sum_{k=1}^{n} {{n}\choose{k}}a^k+a+a\sum_{k=1}^{n} {{n}\choose{k}}a^k}$

$\displaystyle{=\sum_{k=1}^{n} {{n}\choose{k}}a^k+\sum_{k=1}^{n} {{n}\choose{k}}a^{k+1}+a}$

$\displaystyle{=\sum_{k=2}^{n} {{n}\choose{k}}a^k+{{n}\choose{1}}a+{{n}\choose{n}}a^{n+1}+\sum_{k=2}^{n} {{n}\choose{k}}a^k+a}$

$\displaystyle{=\sum_{k=2}^{n} \left[{{n}\choose{k}}+{{n}\choose{k-1}}\right]a^k+{{n+1}\choose{n+1}}a^{n+1}+{{n}\choose{1}}a+a}$

$\displaystyle{={{n+1}\choose{n+1}}a^{n+1}+\sum_{k=2}^{n} {{n+1}\choose{k}}a^k+{{n+1}\choose{1}}a=\sum_{k=1}^{n+1} {{n+1}\choose{k}}a^k}$

To proceed with these equalities we used some basic properties of the binomial coefficient and manipulation of the sums.

Instead of 7, use variable x, then:

• star(x,x) = $x^2 + 2x$
• star(x,star(x,x)) = $x^3 + 3x^2 + 3x$
• star(x,star(x,star(x,x))) = $x^4 + 4x^3 + 6x^2 + 4x$
• ...

These polynomials and the coefficients $1, 4, 6, 4$ look familiar! It's Pascal's triangle, whose rows contain the coefficients for binomial expansions $(a+b)^n$ . To make it fit the triangle's pattern, we choose $a=x, b=1, n=7$ to arrive at $(x + 1)^7 - 1$ .

With $x=7$ , the answer is $8^7 - 1$ .

Matlab solution: