1. Problem statement: Analyze and simplify the expression $1+\sqrt[n]{2}$ for integer $n\ge 1$. 2. Special case and direct evaluation. If $n=1$ then $1+\sqrt[1]{2}=3$. 3. Irrationality for $n\ge 2$. Assume $n\ge 2$ and suppose, for contradiction, that $1+\sqrt[n]{2}$ is rational. Then $\sqrt[n]{2}$ is rational, so write $\sqrt[n]{2}=\frac{p}{q}$ in lowest terms with integers $p,q$ and $q>0$. Raising both sides to the $n$th power gives $p^n=2q^n$. Thus $p^n$ is even, so $p$ is even; write $p=2r$. Substituting back gives $(2r)^n=2q^n$, so $2^{n-1}r^n=q^n$. This implies $q^n$ is even, hence $q$ is even, contradicting that $p/q$ was in lowest terms. Therefore $\sqrt[n]{2}$ is irrational and consequently $1+\sqrt[n]{2}$ is irrational for $n\ge 2$. 4. Algebraic relation and minimal polynomial. Let $x=1+\sqrt[n]{2}$, then $\sqrt[n]{2}=x-1$. Raising to the $n$th power yields the polynomial equation: $$(x-1)^n-2=0$$ This shows $x$ is an algebraic number whose minimal polynomial divides $(x-1)^n-2$. 5. Numerical examples and intuition. If $n=2$ then $1+\sqrt{2}\approx 2.41421356$. If $n=3$ then $1+\sqrt[3]{2}\approx 1+1.25992105=2.25992105$. 6. Final answer. In simplest form the expression is $1+\sqrt[n]{2}$; for $n=1$ it equals 3 and for every integer $n\ge 2$ it is irrational and satisfies $$(x-1)^n-2=0$$.